RANGE OF FUNCTION WORKSHEET

Find the range of the following functions :

Question 1 :

f(x) = (x - 2)/(3 - x)

Question 2 :

f(x) = 1/√(x - 5)

Question 3 :

f(x) = √(16 - x²)

1. Answer :

f(x) = (x - 2)/(3 - x)

Let y = f(x).

y = (x - 2)/(3 - x)

Solve for x in terms of y..

y(3 - x) = (x - 2)

3y - xy = x - 2

3y + 2 - xy = x

3y + 2 = x + xy

3y + 2 = x(1 + y)

x = (3y + 2)/(1 + y)

Now, the equation has been solved for x in terms of y. On the right side of the equation, we have (1 + y) as denominator. If y = -1, the denominator will become zero and the value of x will be undefined.

Hence the range is R - {-1}.

2. Answer :

f(x) = 1/√(x - 5)

Let y = f(x).

y = 1/√(x - 5)

Solve for x in terms of y.

y√(x - 5) = 1

√(x - 5) = 1/y

Squaring both sides,

[√(x - 5)]² = (1/y)²

x - 5 = 1/y²

x = 1/y² + 5

x = (1 + 5y²)/y²

Now, the equation has been solved for x in terms of y. On the right side of the equation, we have y² as denominator. If y = 0, the denominator will become zero and the value of x will be undefined.

In the given function f(x) = 1/√(x - 5), f(x) will take only positive values for any value of x.

Hence the range of f(x) is (0, +∞)

3. Answer :

f(x) = √(16 - x²)

Let y = f(x).

y = √(16 - x²)

Squaring both sides,

y² = 16 - x²

x² = 16 - y²

x = √(16 - y²)

Clearly x will be real, if (16 - y²) ≥ 0

16 - y²) ≥ 0

Multiplying both sides by -1.

y² - 16 ≤ 0

(y + 4)(y - 4) ≤ 0

-4 ≤ y ≤ 4 ----> y ∈ [-4, 4]

Also, y = √(16 - x²) ≥ 0 for all x ∈ [-4, 4].

Thus y ∈ [0, 4] for all x ∈ [-4, 4]

Hence the range of f(x) is [0, 4].

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