A functional relationship between y and x can be made explicit. Even if the relation is not functional, however, we can assume the relation usually defines y as a function.
A functional relationship between y and x can be made explicit. Even if the relation is not functional, however, we can assume the relation usually defines y as a function.
21 - GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES.pptxbernadethvillanueva1
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
We cover the inverses to the trigonometric functions sine, cosine, tangent, cotangent, secant, cosecant, and their derivatives. The remarkable fact is that although these functions and their inverses are transcendental (complicated) functions, the derivatives are algebraic functions. Also, we meet my all-time favorite function: arctan.
21 - GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES.pptxbernadethvillanueva1
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
We cover the inverses to the trigonometric functions sine, cosine, tangent, cotangent, secant, cosecant, and their derivatives. The remarkable fact is that although these functions and their inverses are transcendental (complicated) functions, the derivatives are algebraic functions. Also, we meet my all-time favorite function: arctan.
Fundamentals of AlgebraChu v. NguyenIntegral ExponentsDustiBuckner14
Fundamentals of Algebra
Chu v. Nguyen
Integral Exponents
Exponents
If n is a positive integer (a whole number, i.e., a number without decimal part) and x is a number, then
The number x is called the base and n is called the exponent.
The most common ways of referring to are “ x to the nth power,”
“ x to the nth,” or “the nth power of x.”
Integral Exponents (cont.)
For any non-zero number x and a positive integer n
and
Note: is not defined
and
Rules Concerning Integral Exponents
Following are five rules in which m and n are positive integers:
Rule 1: ; for example,
Rule 2: ; for example
or
Rules Concerning Integral Exponents (Cont.)
Rule 3: ; for example
or
Rule 4: ; for example
or
Rule 5: ; for example
or
Basic Rules for Operating with Fractions
Since dividing by zero is not defined, we assume that the denominator
is not zero.
Following are the eight basic rules for operating with fractions.
Rule 1: ; for example
Rule 2: ; for example
Rule 3: ; for example
Basic Rules for Operating with Fractions (cont.)
Rule 4: ; for example
Rule 5: ; for example
Rule 6: ; for example
Basic Rules for Operating with Fractions (cont.)
Rule 7: ; for example
Rule 8: ; for example
Notes: a*b +a*x may be expressed as a(b + x)
a*b + 1 may be written as a(b + ), and
m*x – y may be expressed as m(x - )
Square Root
Generally, for a>0 , there is exactly one positive number x such that
, we say that x is the root of a, written as
for
When n = 2, we say that x is the square root of “a” and is denoted by
or or
For example:
or
Practices
Carrying out the following operations:
24 ; 2-2 ; 2322, ; 252-5 ; and (2x3)5
; ; ; and
3.
4.
5.
n
m
n
m
x
x
x
+
=
n
x
1
0
)
2
(
2
2
2
=
=
=
-
+
-
x
x
x
x
1
0
=
x
0
0
n
n
n
x
x
x
x
=
=
+
0
0
n
n
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x
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-
3
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This learner's module discusses or talks about the topic of Quadratic Functions. It also discusses what is Quadratic Functions. It also shows how to transform or rewrite the equation f(x)=ax2 + bx + c to f(x)= a(x-h)2 + k. It will also show the different characteristics of Quadratic Functions.
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As a business owner in Delaware, staying on top of your tax obligations is paramount, especially with the annual deadline for Delaware Franchise Tax looming on March 1. One such obligation is the annual Delaware Franchise Tax, which serves as a crucial requirement for maintaining your company’s legal standing within the state. While the prospect of handling tax matters may seem daunting, rest assured that the process can be straightforward with the right guidance. In this comprehensive guide, we’ll walk you through the steps of filing your Delaware Franchise Tax and provide insights to help you navigate the process effectively.
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Looking for professional printing services in Jaipur? Navpack n Print offers high-quality and affordable stationery printing for all your business needs. Stand out with custom stationery designs and fast turnaround times. Contact us today for a quote!
The world of search engine optimization (SEO) is buzzing with discussions after Google confirmed that around 2,500 leaked internal documents related to its Search feature are indeed authentic. The revelation has sparked significant concerns within the SEO community. The leaked documents were initially reported by SEO experts Rand Fishkin and Mike King, igniting widespread analysis and discourse. For More Info:- https://news.arihantwebtech.com/search-disrupted-googles-leaked-documents-rock-the-seo-world/
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Remote sensing and monitoring are changing the mining industry for the better. These are providing innovative solutions to long-standing challenges. Those related to exploration, extraction, and overall environmental management by mining technology companies Odisha. These technologies make use of satellite imaging, aerial photography and sensors to collect data that might be inaccessible or from hazardous locations. With the use of this technology, mining operations are becoming increasingly efficient. Let us gain more insight into the key aspects associated with remote sensing and monitoring when it comes to mining.
2. We use the factored polynomials or rational expressions to determine the signs of the outputs. Sign-Charts and Inequalities
3. We use the factored polynomials or rational expressions to determine the signs of the outputs. That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities
4. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities
5. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities
6. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) Hence, for x = -3/2: (-3/2 – 3)(-3/2 + 1) That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities
7. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) Hence, for x = -3/2: (-3/2 – 3)(-3/2 + 1) is (–)(–) = + . That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities
8. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) Hence, for x = -3/2: (-3/2 – 3)(-3/2 + 1) is (–)(–) = + . And for x = -1/2: (-1/2 – 3)(-1/2 + 1) That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities
9. We use the factored polynomials or rational expressions to determine the signs of the outputs. Example A: Determine the outcome is + or – for x 2 – 2x – 3 if x = -3/2, -1/2. In factored form x 2 – 2x – 3 = (x – 3)(x + 1) Hence, for x = -3/2: (-3/2 – 3)(-3/2 + 1) is (–)(–) = + . And for x = -1/2: (-1/2 – 3)(-1/2 + 1) is (–)(+) = – . That is, given a formula f and a value for x, factor f to determine if the output is positive (f > 0) or negative (f < 0) when f is evaluated with x. Sign-Charts and Inequalities
10. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 Sign-Charts and Inequalities
11. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Sign-Charts and Inequalities
12. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) Sign-Charts and Inequalities
13. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) < 0 Sign-Charts and Inequalities
14. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) < 0 For x = -1/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(+) (–)(+) Sign-Charts and Inequalities
15. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) < 0 For x = -1/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(+) (–)(+) > 0 Sign-Charts and Inequalities
16. Example B. Determine the outcome is + or – for if x = -3/2, -1/2. x 2 – 2x – 3 x 2 + x – 2 x 2 – 2x – 3 x 2 + x – 2 In factored form = (x – 3)(x + 1) (x – 1)(x + 2) Hence, for x = -3/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(–) (–)(+) < 0 For x = -1/2: (x – 3)(x + 1) (x – 1)(x + 2) = (–)(+) (–)(+) > 0 This leads to the sign charts of formulas. The sign- chart of a formula gives the signs of the outputs.
17. Here is an example, the sign chart of f = x – 1: Sign-Charts and Inequalities
18. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 Sign-Charts and Inequalities
19. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The "+" indicates the region where the output is positive i.e. if 1 < x. Sign-Charts and Inequalities
20. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The "+" indicates the region where the output is positive i.e. if 1 < x. Likewise, the "–" indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities
21. Construction of the sign-chart of f. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The "+" indicates the region where the output is positive i.e. if 1 < x. Likewise, the "–" indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities
22. Construction of the sign-chart of f. I. Solve for f = 0 (and denominator = 0 if there is any denominator). Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The "+" indicates the region where the output is positive i.e. if 1 < x. Likewise, the "–" indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities
23. Construction of the sign-chart of f. I. Solve for f = 0 (and denominator = 0 if there is any denominator). II. Draw the real line, mark off the answers from I. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The "+" indicates the region where the output is positive i.e. if 1 < x. Likewise, the "–" indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities
24. Construction of the sign-chart of f. I. Solve for f = 0 (and denominator = 0 if there is any denominator). II. Draw the real line, mark off the answers from I. III. Sample each segment for signs by testing a point in each segment. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The "+" indicates the region where the output is positive i.e. if 1 < x. Likewise, the "–" indicates the region where the output is negative, i.e. x < 1. Sign-Charts and Inequalities
25. Construction of the sign-chart of f. I. Solve for f = 0 (and denominator = 0) if there is any denominator. II. Draw the real line, mark off the answers from I. III. Sample each segment for signs by testing a point in each segment. Here is an example, the sign chart of f = x – 1: 1 f = 0 + + – – – – x – 1 The "+" indicates the region where the output is positive i.e. if 1 < x. Likewise, the "–" indicates the region where the output is negative, i.e. x < 1. Fact : The sign stays the same for x's in between the values from step I (where f = 0 or f is undefined.) Sign-Charts and Inequalities
26. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Sign-Charts and Inequalities
27. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Sign-Charts and Inequalities
28. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) 4 -1 Sign-Charts and Inequalities
29. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) Select points to sample in each segment: 4 -1 Sign-Charts and Inequalities
30. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) 4 -1 Select points to sample in each segment: Test x = - 2, -2 Sign-Charts and Inequalities
31. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . -2 Sign-Charts and Inequalities
32. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Sign-Charts and Inequalities
33. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Sign-Charts and Inequalities
34. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Sign-Charts and Inequalities
35. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Sign-Charts and Inequalities
36. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + – – – – – 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Sign-Charts and Inequalities
37. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + – – – – – 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Test x = 5, get + * + = +. 5 Sign-Charts and Inequalities
38. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + – – – – – 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Test x = 5, get + * + = +. Hence this segment is positive. Put + over it. 5 Sign-Charts and Inequalities
39. Example C. Let f = x 2 – 3x – 4 , use the sign - chart to indicate when is f = 0, f > 0, and f < 0 . Solve x 2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 , -1 Mark off these points on a line: (x-4)(x+1) + + + + + – – – – – + + + + + 0 4 -1 Select points to sample in each segment: Test x = - 2, get – * – = + . Hence the segment is positive. Draw + sign over it. -2 Test x = 0, get – * + = –. Hence this segment is negative. Put – over it. Test x = 5, get + * + = +. Hence this segment is positive. Put + over it. 5 Sign-Charts and Inequalities
40. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) Sign-Charts and Inequalities
41. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. Sign-Charts and Inequalities
42. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Sign-Charts and Inequalities
43. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. Sign-Charts and Inequalities
44. Example D. Make the sign chart of f = (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 Sign-Charts and Inequalities
45. Example D. Make the sign chart of f = Select a point to sample in each segment: (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 0 2 4 Sign-Charts and Inequalities
46. Example D. Make the sign chart of f = Select a point to sample in each segment: Test x = -3, we get a (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 ( – ) ( – )( – ) = – segment. 0 2 4 Sign-Charts and Inequalities
47. Example D. Make the sign chart of f = Select a point to sample in each segment: Test x = -3, we get a (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 ( – ) ( – )( – ) = – segment. 0 2 4 Test x = 0, we get a ( – ) ( – )( + ) = + segment. Sign-Charts and Inequalities
48. Example D. Make the sign chart of f = Select a point to sample in each segment: Test x = -3, we get a (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 ( – ) ( – )( – ) = – segment. 0 2 4 Test x = 0, we get a ( – ) ( – )( + ) = + segment. Test x = 2, we get a ( – ) ( + )( + ) segment. = – Sign-Charts and Inequalities
49. Example D. Make the sign chart of f = Select a point to sample in each segment: Test x = -3, we get a (x – 3) (x – 1)(x + 2) The root for f = 0 is from the zero of the numerator which is x = 3. The zeroes of the denominator x = 1, -2 are the values where f is undefined (UDF). Mark these values on a real line. (x – 3) (x – 1)(x + 2) -2 1 3 UDF UDF f=0 -3 ( – ) ( – )( – ) = – segment. 0 2 4 Test x = 0, we get a ( – ) ( – )( + ) = + segment. Test x = 2, we get a ( – ) ( + )( + ) segment. = – Test x = 4, we get a ( + ) ( + )( + ) segment. = + – – – – + + + – – – + + + + Sign-Charts and Inequalities
50. The easiest way to solve a polynomial or rational inequality is use the sign-chart. Sign-Charts and Inequalities
51. The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, Sign-Charts and Inequalities
52. The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, Sign-Charts and Inequalities
53. The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Sign-Charts and Inequalities
54. Example E. Solve x 2 – 3x > 4 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Sign-Charts and Inequalities
55. Example E. Solve x 2 – 3x > 4 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Set one side to 0, we get x 2 – 3x – 4 > 0; Sign-Charts and Inequalities
56. Example E. Solve x 2 – 3x > 4 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. Sign-Charts and Inequalities
57. Example E. Solve x 2 – 3x > 4 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities
58. Example E. Solve x 2 – 3x > 4 4 -1 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, (x – 4)(x + 1) Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities
59. Example E. Solve x 2 – 3x > 4 0 4 -1 -2 5 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities
60. Example E. Solve x 2 – 3x > 4 0 4 -1 -2 5 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities
61. Example E. Solve x 2 – 3x > 4 0 4 -1 -2 5 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + – – – – – – Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities
62. Example E. Solve x 2 – 3x > 4 0 4 -1 -2 5 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + – – – – – – + + + + Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities
63. Example E. Solve x 2 – 3x > 4 0 4 -1 The solutions the positive region, {x < -1} U {4 < x} -2 5 4 -1 The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + – – – – – – + + + + Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities
64. Example E. Solve x 2 – 3x > 4 0 4 -1 The solutions the positive region, {x < -1} U {4 < x} -2 5 4 -1 Note: The empty dot means those numbers are excluded. The easiest way to solve a polynomial or rational inequality is use the sign-chart. To do this, I. set one side of the inequality to 0, II. factor the expression and draw the sign-chart, III. read off the answer from the sign chart. Draw the sign-chart, sample the points x = -2, 0, 5 (x – 4)(x + 1) + + + – – – – – – + + + + Set one side to 0, we get x 2 – 3x – 4 > 0; factor, we've (x – 4)(x + 1) > 0. The roots are x = -1, 4. Sign-Charts and Inequalities
65. Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation.
66. Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form,
67. Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction,
68. Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.
69. Example E. Use the sign-chart to solve x – 2 4 < 2 Sign-Charts and Inequalities For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.
70. Example E. Use the sign-chart to solve Set one side to 0, put the expression in the factored form x – 2 4 < 2 Sign-Charts and Inequalities x – 2 4 < – 2 0 For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.
71. Example E. Use the sign-chart to solve Set one side to 0, put the expression in the factored form x – 2 4 < 2 Sign-Charts and Inequalities x – 2 4 < – 2 0 x – 2 4 < – 0 x – 2 2(x – 2) For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.
72. Example E. Use the sign-chart to solve Set one side to 0, put the expression in the factored form x – 2 4 < 2 Sign-Charts and Inequalities x – 2 4 < – 2 0 x – 2 4 < – 0 x – 2 2(x – 2) x – 2 4 – 2(x – 2) < 0 For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.
73. Example E. Use the sign-chart to solve Set one side to 0, put the expression in the factored form x – 2 4 < 2 Sign-Charts and Inequalities x – 2 4 < – 2 0 x – 2 4 < – 0 x – 2 2(x – 2) x – 2 4 – 2(x – 2) < 0 x – 2 8 – 2x < 0 For a fractional inequality, we may not clear the denominator as in the case for equation. Instead we set one side of the inequality to be 0 and put the expression in factored form, i.e. combine into one fraction, then do the sign-chart of the expression.
74. Hence the problem is transformed to Sign-Charts and Inequalities x – 2 8 – 2x < 0
75. Hence the problem is transformed to Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x
76. Hence the problem is transformed to Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2,
77. Hence the problem is transformed to 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence x – 2 8 – 2x 2 UDF
78. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 0, we get a ( + ) ( – ) segment. = – x – 2 8 – 2x 2 UDF
79. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 0, we get a ( + ) ( – ) segment. = – – – – – – – x – 2 8 – 2x 2 UDF
80. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 – – – – – – x – 2 8 – 2x 2 UDF
81. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 + + + – – – – – – x – 2 8 – 2x 2 UDF
82. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 5 Test x = 5, we get a ( + ) segment. ( – ) = – + + + – – – – – – x – 2 8 – 2x 2 UDF
83. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 5 Test x = 5, we get a ( + ) segment. ( – ) = – + + + – – – – – – – – – – – x – 2 8 – 2x 2 UDF
84. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 5 Test x = 5, we get a ( + ) segment. ( – ) = – + + + – – – – – – – – – – – x – 2 8 – 2x 2 UDF The answers are the negative portions as shown,
85. Hence the problem is transformed to 0 4 Sign-Charts and Inequalities x – 2 8 – 2x < 0 To solve it we do the sign-chart of . x – 2 8 – 2x The roots from the numerator is x = 4 and the root from the denominator is 2, hence Test x = 3, we get a ( + ) ( + ) = + segment. Test x = 0, we get a ( + ) ( – ) segment. = – 3 5 Test x = 5, we get a ( + ) segment. ( – ) = – + + + – – – – – – – – – – – x – 2 8 – 2x 2 UDF The answers are the negative portions as shown, or that {x < –2} U {4 ≤ x }.
86. Example F. Solve x – 2 2 < x – 1 3 Sign-Charts and Inequalities
87. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Sign-Charts and Inequalities
88. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, Sign-Charts and Inequalities
89. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) Sign-Charts and Inequalities
90. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities
91. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Sign-Charts and Inequalities
92. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 It has root at x = 4, and its undefined at x = 1, 2. Sign-Charts and Inequalities
93. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. Sign-Charts and Inequalities
94. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 2 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities
95. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities
96. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities
97. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + – – 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities
98. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + – – + + + + 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities
99. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + – – + + + + – – – – 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 Sign-Charts and Inequalities
100. Example F. Solve x – 2 2 < x – 1 3 Set the inequality to 0, x – 2 2 x – 1 3 < 0 Put the expression into factored form, x – 2 2 x – 1 3 = (x – 2)(x – 1) 2(x – 1) – 3(x – 2) = (x – 2)(x – 1) – x + 4 Hence the inequality is (x – 2)(x – 1) – x + 4 < 0 Draw the sign chart by sampling x = 0, 3/2, 3, 5 It has root at x = 4, and its undefined at x = 1, 2. 4 1 0 5 + + + – – + + + + – – – – 2 3/2 3 UDF UDF (x – 2)(x – 1) – x + 4 We want the shaded negative region, i.e. {1 < x < 2} U {4 < x}. Sign-Charts and Inequalities
101. Sign-Charts and Inequalities Exercise A. Draw the sign–charts of the following formulas. 1. (x – 2)(x + 3) 4. (2 – x)(x + 3) 5. –x(x + 3) 7. (x + 3) 2 9. x(2x – 1)(3 – x) 12. x 2 (2x – 1) 2 (3 – x) 13. x 2 (2x – 1) 2 (3 – x) 2 14. x 2 – 2x – 3 16. 1 – 15. x 4 – 2x 3 – 3x 2 (x – 2) (x + 3) 2. (2 – x) (x + 3) 3. – x (x + 3) 6. 8. –4(x + 3) 4 x (3 – x)(2x – 1) 10. 11. x 2 (2x – 1)(3 – x) 1 x + 3 17. 2 – 2 x – 2 18. 1 2x + 1 19. – 1 x + 3 – 1 2 x – 2 20. – 2 x – 4 1 x + 2
102. Sign-Charts and Inequalities Exercise B. Use the sign–charts method to solve the following inequalities. 21. (x – 2)(x + 3) > 0 23. (2 – x)(x + 3) ≥ 0 28. x 2 (2x – 1) 2 (3 – x) ≤ 0 29. x 2 – 2x < 3 33. 1 < 32. x 4 > 4x 2 (2 – x) (x + 3) 22. – x (x + 3) 24. 27. x 2 (2x – 1)(3 – x) ≥ 0 1 x 34. 2 2 x – 2 35. 1 x + 3 2 x – 2 36. > 2 x – 4 1 x + 2 25. x(x – 2)(x + 3) x (x – 2)(x + 3) 26. ≥ 0 30. x 2 + 2x > 8 30. x 3 – 2x 2 < 3x 31. 2x 3 < x 2 + 6x ≥ ≥ 0 ≤ 0 0 ≤ 37. 1 < 1 x 2