FUNCTIONS EVALUATION

1. Evaluating Functions and
Operations on Functions

2. Function Notation
y = x2 – 4 f(x) = x2 – 4
•Pronounced f of x
Identifies the equation as a
function Indicates which variable the
function is in terms of (the
variable used in the function)

5. Evaluating Function
Remember that:
f(x) = 3x +1:
1. f(x) means “the value of f at x”. It does not
mean “f times x.”
2.Letters other than f such as G and H or g
and h can also be used.
3. f is the name of the function and f(x) is the
value of the function at x.

8. Evaluating Functions Algebraically
•Solve for the value of a function at a
point.
•f(x) = 2x – 10; find f(6)
•f(6) = 2(6) – 10
f(6) = 12 – 10
f(6) = 2
The value of x is 6.
Replace x with 6 and
solve.

9. Function Notation
9
Given g(x) = x2 – 3, find g(-2) .
g(-2) = x2 – 3
g(-2) = (-2)2 – 3
g(-2) = 1

10. Function Notation
10
Given f(x) = , the following.
f(3) = 2x2 – 3x
f(3) = 2(3)2 – 3(3)
f(3) = 2(9) - 9
f(3) = 9
a. f(3) b. 3f(x) c. f(3x)
3f(x) = 3(2x2 – 3x)
3f(x) = 6x2 – 9x
f(3x) = 2x2 – 3x
f(3x) = 2(3x)2 – 3(3x)
f(3x) = 2(9x2) – 3(3x)
f(3x) = 18x2 – 9x
2
2 3
x x


11. If f(x) = x + 12, evaluate each
a. f (4) b. f (-2) c. f (-x) d. f (x+3)
a. Replacing x with 4
f(4) = 4 + 12 = 16
b. Replacing x with -2
f(-2) = -2 + 12 = 10
c. Replacing x with -x
f(-x) = -x + 12
d. Replacing x with x+3
f(x+3) = x + 3 + 12
f(x + 3) = x + 15

12. 6(-5) + 100
-30 + 100
70

13. h(4) = -3(4) -5
h(4) = -12 - 5
h(4) = -17
h(4) = -3x -5
h(-3) = -3(-3) -5
h(-3) = 9- 5
h(-3) = 4
h(-3) = -3x -5

14. Evaluating Functions Algebraically,
cont.
•You can also evaluate functions with
an expression
•f(x) = 2x – 10; find f(x+1)
f(x+1) = 2(x+1) – 10
f(x+1) = 2x + 2 – 10
f(x+1) = 2x – 8

15. f(-1) = (-1)2 +3(-1) -4
f(-1) = 1 – 3 - 4
f(-1) = -6
f(-1) = x2 +3x-4
h(4) = (4)2 +3(4) -4
h(4) = 16 + 12 - 4
h(4) = 24
h(4) = x2 +3x-4

16. EXPONENTS
f(6) = (-12 +6 )2
f(6) = (-12 ) 2 + (6 )2
f(6) = 144 + 36
f(6) = (-12 +x)2
f(6) = 180

17. INTEGERS
g(-4) = -3 + (-4)
g(-4) = -3 - 4
g(-4) = -7
g(-4) = -3 + x

18. f(-10) = (-1)2 +3(-1) -4
f(-10) = 1 – 3 - 4
f(-10) = -6
f(-10) = x2 +3x-4

19. f(-1) = (-1)2 +3(-1) -4
f(-1) = 1 – 3 - 4
f(-1) = -6
f(-1) = x2 +3x-4

20. f(-1) = (-1)2 +3(-1) -4
f(-1) = 1 – 3 - 4
f(-1) = -6
f(-1) = x2 +3x-4