# CLASS-11RELATION & FUNCTION - FURTHER DISCUSSION OF RANGE OF FUNCTION

Further Discussion on the Range of a Function

Once we know the domain of the function, we can determine the values f(x) can take. The set of all these values is called the range of the function.

Method for finding the range of a function y = f(x)

Step.1) Find the domain of the function y = f(x)

Step.2) Solve the equation y = f(x) to find x in term of y. Find the real values of y for which x is real.

The set of values of y so obtained makes up the range of y.

For Example,

1.) The function f(x) = x(8 – x) can be evaluated for any real number x, so the domain is the complete set of real numbers.

Now, let y = f(x)

=> y = x (8 – x)

=> x²- 8x + y = 0

(- 8) ± √{(- 8)² - 4 X 1 X y}

=>  x = -  -----------------------------

2 X 1

8 ± √{(- 8)²- 4 X 1 X y}

=>   x  = --------------------------

2

8 ± √(64 – 4y)

=>   x =  -----------------

2

x will be real if 64 – 4y ≥ 0

=>  64 ≥ 4y

=>   16 ≥ y   or   y ≤ 16

So, the range consists of all real numbers less than or equal to 16. (Ans.)

2.) Let y = f(x) = √x (8 – x)

Its domain is the set of real numbers 0 ≤ x ≤ 8.

Now, y = √8x – x²

=>  y² = 8x - x²

=>  x² - 8x + y² = 0

8 ± √64 – 4y²

=>  x = --------------

2

x will be real if 64 – 4y² ≥ 0

=>  y² - 16 ≤ 0

=>  (y + 4) (y – 4) ≤ 0

The above inequality is true when – 4 ≤ y ≤ 4

=>  y ∈ [- 4, 4]

Now, y = √16 - x² ≥ 0 for all x ∈ [- 4, 4]

So,  y ∈ [0, 4] to all x ∈ [- 4, 4]

Hence, range (f) = [0, 4]               (Ans.)

Example.1) Find the range of the following function.

(i)  y =  x²/(1 + x²)

x²

Ans.) Since,  y = -------- is defined for all real x, therefore,

1 + x²

domain of y = (- ∞, ∞)

x²

Now, y = ---------

1 + x²

=>  y + x²y = x²

=>  y = x² - x²y

=>  x² (1 – y) =  y

y

=>  x² = -------

1 – y

y

=>  x = √-------

1 – y

y            1 – y          y(1 – y)

=>  x = √-------- . --------- = √-----------

1 – y         1 – y           (1 – y)²

√y (1 – y)

=>  x = ------------

1 – y

For x to be real, 1 – y ≠ 0, i.e., y  ≠ 1 and y (1 – y) ≥ 0

=>  y(y – 1) ≤ 0 and y ≠ 1

=>  0 ≤ y < 1.

Range of y = [0, 1)           (Ans.)

(ii) f(x) = √(3x²- 4x + 5) ≥ 0

Ans.)  f(x) is defined if 3x²- 4x + 5

4            5

=>  3 (x² - ------ x + ------) ≥ 0

3            3

4            4          4          5

=>  3 (x²- ------ x + ------ - ------ + ------) ≥ 0

3            9          9          3

2          (15 – 4)

=>  3[(x - -----)² + ----------] ≥ 0

3             9

2          11

=>  3[(x - -----)² + -----] ≥ 0

3           9

Which is true for all real x.

So, Domain (f) = (- ∞, ∞)

Let y = √3x²- 4x + 5

=>  y² = 3x²- 4x + 5

=>  3x²- 4x + 5 - y² = 0

For x to be real, (- 4)²- (4 X 3) X (5 - y²) ≥ 0

=>  y  ≥  √(11/3)

So, the range of y = [√11/3, ∞)              (Ans.)

(iii) f(x) = logₑ (3x²- 4x + 5)

Ans.)  f(x) is defined if 3x²- 4x + 5 > 0

4            5

=>  3 (x² - ------ x + ------) > 0

3            3

4            4          4          5

=>  3 (x² - ------ x + ------ - ------ + ------) > 0

3            9          9          3

2          (15 – 4)

=>  3[(x - -----)² + ----------] > 0

3             9

2           11

=>  3[(x - ------)² + ------- ] > 0 which is the true for all real x

3            9

So, the Domain of f = (- ∞, ∞).

Let, y = logₑ (3x² - 4x + 5)

=>  ₑy = 3x² - 4x + 5

=> 3x² - 4x + 5 - ₑy = 0

For x to be real,

(- 4)²- 12 (5 - ₑy) ≥ 0

=> 12 ₑy ≥ 44

=>  ₑy ≥ 11/3

=>  y ≥ logₑ  11/3

11

So, the Range of f = [logₑ -----, ∞)       (Ans.)

3

(iv) f(x) =  5 / (3 - x²)

Ans.)  For f(x) to be defined, 3 - x² ≠ 0, i.e., x ≠ ± √3

So, Domain of f = R – {- √3, √3}

5

Now, let  y = --------

3 - x²

=>  3y - x²y = 5

=>  x²y = 3y – 5

(3y – 5)

=>  x² = ----------

y

(3y – 5)

=>    x  =  √----------

Y

Now, x will take real values other than - √3 and √3, if

3y - 5

=>   --------- ≥ 0

y

5

=>    y < 0  and  y ≥ ------. Also, y ≠ 0        (Ans.)

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