**Finding Inverse of a Function for Particular Value :**

Here we are going to see, how we find inverse of a function for a particular value.

**Question 1 :**

Suppose f(x) = x^{2} − 1, with the domain of f being the set of positive numbers.

(a) Evaluate f^{−1}(8).

(b) Evaluate f(8)^{−1}.

**Solution :**

f(x) = x^{2} − 1

y = f(x)

y = x^{2} − 1

x^{2} = y + 1

x = √(y + 1)----(1)

x = f^{-1}(y) ----(2)

(1) = (2)

f^{−1}(8) = √(8 + 1)

f^{−1}(8) = 3

(b) f(8)^{−1} = [1/f(8)]

= 1/(8^{2} − 1)

= 1/63

**Question 2 :**

Suppose f(x) = 4x + 6. Evaluate f^{−1}(5).

**Solution :**

y = f(x)

4x + 6 = y

x = (y - 6)/4 ----(1)

x = f^{-1}(y) ----(2)

(1) = (2)

f^{−1}(5) = (5 - 6)/4

f^{−1}(5) = -1/4

**Question 3 :**

Suppose f(x) = 7x − 5. Evaluate f^{−1}(−3).

y = f(x)

7x - 5 = y

x = (y + 5)/7 ----(1)

x = f^{-1}(y) ----(2)

(1) = (2)

f^{−1}(-3) = (-3 + 5)/7

f^{−1}(-3) = 2/7

**Question 4 :**

Suppose f(x) = 2 + [(x − 5)/(x + 6)].

(a) Evaluate f ^{−1}(4).

(b) Evaluate f(4)^{−1}.

**Solution :**

y = f(x)

y = 2 + [(x − 5)/(x + 6)]

y(x + 6) = 2(x + 6) + (x - 5)

xy + 6y = 2x + 12 + x - 5

xy + 6y = 3x + 7

3x - xy = 6y - 7

x(3 - y) = (6y - 7)

x = (6y - 7)/(3 - y) -----(1)

x = f^{-1}(y) ----(2)

(1) = (2)

f^{−1}(4) = (6(4) - 7)/(3 - 4)

f^{−1}(4) = (24 - 7)/(-1)

f^{−1}(4) = -17

(b) Evaluate f(4)^{−1}.

f(x) = 2 + [(x − 5)/(x + 6)]

f(4) = 2 + [(4 − 5)/(4 + 6)]

f(4) = 2 + (-1/10)

f(4) = (20 - 1)/10

f(4)^{−1 } = 1/(19/10)

f(4)^{−1 } = 10/19

**Question 5 :**

Suppose h(x) = 3x^{2} − 4, where the domain of h is the set of positive numbers. Find a formula for h^{−1}

**Solution :**

y = h(x)

Given that : h(x) = 3x^{2} − 4

y = 3x^{2} − 4

(y + 4) = 3x^{2}

x^{2} = (y + 4)/3

x = √(y + 4)/3

h^{−1 }= √(y + 4)/3

After having gone through the stuff given above, we hope that the students would have understood "Finding Inverse of a Function for Particular Value".

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