HOW TO FIND INVERSE OF A FUNCTION

Example 1 :

Find the inverse of the function f(x) = x - 5.

Solution :

f(x) = x - 5

Replace f(x) by y.

y = x - 5

Interchange x and y. 

x = y - 5

Solve for y.

y = x + 5

Replace y by f^-1(x).

f^-1(x) = x + 5

Example 2 :

Find the inverse of the function f(x) = 3x + 5.

Solution :

f(x) = 3x + 5

Replace f(x) by y.

y = 3x + 5

Interchange x and y. 

x = 3y + 5

Solve for y.

x - 5 = 3y

y = (x - 5)/3

Replace y by f^-1(x).

f^-1(x) = (x - 5)/3

f^-1 (x)  =  (x - 5)/3

Example 3 :

Find the inverse of the function f(x) = x².

Solution :

Replace f(x) by y.

y = x²

Interchange x and y. 

x = y²

y² = x

Solve for y.

Take square root on both sides. 

y = ±√x

Replace y by f^-1(x).

f^-1(x) = ±√x

Example 4 :

Find the inverse of the function f(x) = log₅(x). 

Solution :

f(x) = log₅(x)

Replace f(x) by y.

y = log₅(x)

Interchange x and y. 

x = log₅(y)

Solve for y.

y = 5^x

Replace y by f^-1(x).

f^-1(x) = 5^x

Example 5 :

Find the inverse of the function f(x) = √(x + 1). 

Solution :

f(x) = √(x + 1)

Replace f(x) by y.

y = √(x + 1)

Interchange x and y. 

x = √(y + 1)

Solve for y.

x² = y + 1

y = x² - 1

Replace y by f^-1(x).

f^-1(x) = x² - 1

Example 6 :

Find the inverse of the function f(x) = (x + 2)/(x - 5). 

Solution :

f(x) = (x + 2)/(x - 5)

Replace f(x) by y.

y = (x + 2)/(x - 5)

Interchange x and y. 

x = (y + 2)/(y - 5)

Solve for y.

x(y - 5) = y + 2

xy - 5x = y + 2

xy - y = 5x + 2

y(x - 1) = 5x + 2

y = (5x + 2)/(x - 1)

Replace y by f^-1(x).

f^-1(x) = (5x + 2)/(x - 1)

Example 7 :

Consider the function f(x) = 2x³ + 1. Determine whether the inverse of f is a function. Then find the inverse.

Solution :

y = 2x³ + 1

Finding inverse :

y - 1 = 2x³

x³ = (y - 1)/2

x = ∛(y - 1)/2

Put x = y and y = x

y = ∛(x - 1)/2

so, inverse of the given function is y = ∛(x - 1)/2.

Example 8 :

Consider the function f(x) = 2√(x − 3). Determine whether the inverse of f is a function. Then find the inverse

Solution :

f(x) = 2√(x − 3)

Let y = 2√(x − 3)

y/2 = √(x − 3)

Squaring on both both sides

(y/2)² = x - 3

y²/4 = x - 3

x = (y²/4) + 3

Change x = y and y = x

y = (x²/4) + 3

f^-1(x) = (x²/4) + 3

Example 9 :

Find the inverse of the function that represents the surface area of a sphere, 

S = 4πr²

Then find the radius of a sphere that has a surface area of 100π square feet.

Solution :

S = 4πr²

Divide by 4π, we get

r² = S/4π

r = √(S/4π)

Let us apply S = 100π square feet to find the value of r.

r = √(100π/4π)

r = √(100/4)

r = √25

= 5 feet

So, the radius is 5 feet.

Example 10 :

The distance d (in meters) that a dropped object falls in t seconds on Earth is represented by

d = 4.9t²

Find the inverse of the function. How long does it take an object to fall 50 meters?

Solution :

d = 4.9t²

To find inverse, we divide 4.9 on both sides.

t² = d/4.9

t = √(d/4.9)

When d = 50 meters, t = ?

t = √(50/4.9)

t = √10.20

t = 3.19 seconds

Example 11 :

The maximum hull speed v (in knots) of a boat with a displacement hull can be approximated by v = 1.34 √ℓ , whereℓ is the waterline length (in feet) of the boat. Find the inverse function. What waterline length is needed to achieve a maximum speed of 7.5 knots?

Solution :

v = 1.34 √ℓ

Divide by 1.34, we get

v/1.34 = √ℓ

Squaring on both sides.

l = (v/1.34)²

l = v² / 1.7956

When v = 7.5 knots, l = ?

l = (7.5)² / 1.7956

l = 56.25/1.7956

l = 31.32 feet

Example 12 :

Elastic bands can be used for exercising to provide a range of resistance. The resistance R (in pounds) of a band can be modeled by

R = (3/8) L − 5

where L is the total length (in inches) of the stretched band. Find the inverse function. What length of the stretched band provides 19 pounds of resistance?

Solution :

R = (3/8) L − 5

R + 5 = (3/8) L

L = (8/3)(R + 5)

When R = 19

L = (8/3)(19 + 5)

= (8/3)(24)

= 8(8)

= 64

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