The document explores various functions and their domains by presenting multiple examples, each demonstrating how to determine when a function is defined based on its algebraic expressions. It discusses various functions such as rational, radical, logarithmic, and absolute functions while emphasizing the necessary conditions for their validity. Additionally, it includes solutions and methodologies for evaluating the domain of real functions, showcasing critical points of interest where the functions are undefined.

1. FUNCTIONS
FUNCTIONS

2. FUNCTIONS
PROBLEMS BASED ON
DOMAIN & RANGE

3. FUNCTIONS
1. Find the domain of the real function f(x) =
1
(x2 − 1)(x + 3)
f(x) is not defined when,
(x2-1)(x+3)=0
(x-1)(x+1)(x+3)=0
x =1,-1,-3
domain of f is R-{1, -1, -3}
Solution
(x-1)=0 or (x+1)=0 or (x+3)=0
Given function is
in the form of
1
f(x)
Here, what is
the domain of
f(x)?

4. FUNCTIONS
2. Find the domain of the real function
2x2 − 5x + 7
(x − 1) (x − 2) (x − 3)
Here f(x) is not defined when,
(x-1)(x-2)(x-3)=0
Let f(x) =
2x2 − 5x + 7
(x−1)(x−2)(x−3)
Solution
(x-1)=0 or (x -2)=0 or (x-3) = 0
Where f(x) is not
defined?
 x = 1,2,3
domain of f is R-{1, 2, 3}
Guess the
domain of f(x).

5. FUNCTIONS
3. Find the domain of the real function f(x)= 16 − x𝟐
Domain of f(x) is [-4, 4]
16−x𝟐 is defined when,
x2 -16  0
x  [-4, 4]
Solution
(x + 4) (x - 4)  0
16 - x2  0
When f(x) is
defined?
If
(x-) (x-)≤0x[,]
(x-(-4)(x - 4)  0

6. FUNCTIONS
Domain of f(x) is (-, -5][5, )
4. Find the domain of the real function f(x) = x𝟐 − 25
x𝟐 − 25 is defined when,
(x-5)(x+5)  0
x(-, -5][5, )
Solution
x𝟐
− 25  0
x-5 x  5
(or)
When f(x) is
defined?
(x-5)(x-(-5))  0
If
(x-) (x-)≥0x(-, ][, )
x≤ or x≥ 

7. FUNCTIONS
5. Find the domain of the real function f(x) = 4x − x𝟐
Domain of f(x) is [0,4]
4x − x𝟐 is defined when,
x𝟐
-4x  0
x[0,4]
(x-0)(x-4)  0
Solution
4x − x𝟐
 0
x(x-4)  0
0x 4
When f(x) is
defined?
If
(x-) (x-)≤0x[,]

8. FUNCTIONS
6. Find the domain of the real function f(x) = 2−x + 1+x
x  -1(2)
2−x is defined when,
x - 2  0
1+x is defined when,
Solution
2 - x  0
x  2(1)
1 + x  0
Here we find
domain of
2−x , 1+x
seperately
When 2−x is
defined?
When 1+x is
defined?
-2 -1 0 1 2 
-
Hence from (1)
-1  x  2 x  [-1, 2]
and (2)

9. FUNCTIONS
7. Find the domain of the real function f(x) =
1
1 − x𝟐
f(x) is defined when,
x2 – 1 < 0
Solution
(x +1) (x -1) < 0
1 - x2 > 0
(x –(-1)) (x -1) < 0
Domain of f(x) is (-1,1)
x(-1, 1)
-1 < x < 1
If
(x-) (x-)<0x(,)

10. FUNCTIONS
8. Find the domain of the real function log(x2 - 4x + 3)
Let f(x)=log(x2 -4x + 3)
f(x) is defined when
(x - 1) (x - 3) > 0
Domain of f(x) is (-, 1)(3, )
x (-, 1)(3, )
Solution
x2 - 4x + 3 > 0
If (x-) (x-)>0x(-,)(,)
x< or x>

11. FUNCTIONS
9. Find the domain of the real function f(x) =
1
log(2 − x)
2 - x > 0 and 2 - x  1
x - 2 < 0 and 2 -1  x
f(x) =
1
log(2 − x)
is defined when
Solution :
x < 2 and x  1
2
1
0
-1
-2
×
-∞ +∞
(OR) x(-∞,1)(1,2)
x (-, 2)-{1}

12. FUNCTIONS
10. Find the domain of the real function
2 + x + 2 − x
x
Let f(x) =
2 + x + 2 − x
x
2 + x is defined when
x + 2  0
2−x is defined when
Solution
2 + x  0
x  -2  (1)
2 - x  0
x - 2  0
x  2  (2)

13. FUNCTIONS
2
1
0
-1
-2
×
-∞ +∞
From (1), (2) and (3)
1
x
is defined when x  0(3)
x  -2  (1)
x  2  (2)
 x  [-2, 0)  (0, 2]

14. FUNCTIONS
11.Find the domain of the real valued function
f(x)= log0.3(x − x𝟐)
log0.3(x−x𝟐) is defined when
Solution 0 < (x - x2)  1
(x - x2) >0 and (x - x2)  1
i) x - x2 > 0
x2 - x < 0
x (x - 1) < 0
(x-0)(x-1)<0
 x  (0, 1)

15. FUNCTIONS
Also ii) (x - x2)  1
x2 – x +1  0
x −
1
2
𝟐
+
3
4
 0 (True  x R)
 domain = (0, 1)  R = (0, 1)

16. FUNCTIONS
12. Find the domain f(x)= x − [x]
Domain of f(x) is R
x − [x] is defined when
Solution
x - [x]  0 
which is true  x  R
x  [x]
When f(x)
is defined?

17. FUNCTIONS
13. Find the domain of f(x)= [x]−x
But [x] > x is not possible.
f(x) exists when
[x] > x (or)
Now [x] = x
Domain of f(x) is Z
Solution
[x] - x  0
[x]  x
[x] = x
 x  Z
So, we must have [x] = x
Is it
possible?

18. FUNCTIONS
14. Find the domain f(x) = x − x
Domain of f(x) is R
x − x is defined when
Solution
x − x  0 which is true  x  R

19. FUNCTIONS
15. Find the domain of f(x) = log(x - [x]) ?
 Domain of f(x) is R - Z.
log(x - [x]) is defined when
Solution
x - [x] > 0
x > [x] which is true  x R - Z.

20. FUNCTIONS
16. Find the domain and range of f(x) =
x
2 − 3x
f(x) =
x
2 − 3x
does not exists when,
 Domain of f is R -
2
3
Let y = f(x)
Solution
2 - 3x = 0
x =
2
3
 y =
x
2 − 3x
When is
f(x) not
defined?

21. FUNCTIONS
2y - 3xy = x
x =
2y
1 + 3y
Range of f is R -
−1
3
y (2 - 3x) = x
2y = x (1 + 3y)
which is defined when, 1 + 3y  0
y 
−1
3
2y = x + 3xy

22. FUNCTIONS
17. Find the domain and range of the f(x)= 9−x𝟐
Domain of f(x) is [-3, 3]
(i) Domain
f(x)= 9 − x𝟐 is defined when,
x2 – 9  0
x  [-3, 3]
Solution
9 - x2  0
(x + 3) (x - 3)  0
-3  x  3
(x – (-3)) (x - 3)  0

23. FUNCTIONS
(ii) Range
We know that x2  0
The range of the f(x) is [0, 3]
-x2  0
9 – x2  9
9 − x2  3
and We know that 9−x2  0
𝟎 9−x2 (2)
From (1) & (2), 𝟎  9 − x2  3
f(x)= 9−x𝟐 0,  x  [-3,3]
(1)

24. FUNCTIONS
18. Find the range of the real function
x𝟐
− 4
x − 2
Let f ( x ) =
x𝟐
− 4
x − 2
Solution
f ( x ) does not exist when x – 2 = 0
x = 2
Domain = R - { 2 }
Also f(x) =
x𝟐
− 4
x − 2
= x + 2
=
(x + 2) (x − 2)
(x − 2)
When is
f(x) not
defined?

25. FUNCTIONS
Now f ( x ) = x + 2
Thus f ( x ) can take any value except 4.
Range = R - { 4 }.
But x  2
x + 2  2 + 2
f ( x )  4

26. FUNCTIONS
19. Find the domain and range of the real valued function
f(x)=
2+x
2−x
Domain is R-{2}
Let f ( x ) =
2 + x
2 − x
Let y =
2 + x
2 − x
Solution
2 – x = 0
f ( x ) does not exist when,
x = 2
y ( 2 – x ) = 2 + x

27. FUNCTIONS
y +1  0
Range is R - {-1}
2 y – x y = 2 + x
 x =
2 y − 2
y + 1
2 y – 2 = x + x y
2 y - 2 = x ( 1 +y )
which is defined when,
y  -1
2 y − 2
y + 1
= x

28. FUNCTIONS
20.Find the domain and range of the function
f(x)=|x|+|1+x|
f(x) =|x|+|1 + x| is defined
Domain is R.
We know u + v  u + v for u, v  R
Put u = -x , v = 1 + x in above equation
x + 𝟏 + x  1
f(x)  1
Solution
1  −x + 𝟏 + x
−x + 1 + x  −x + 1 + x
( ∵ −x = x )
Range = [1, ∞)

29. FUNCTIONS
Thank you…