Question

Which of the following pairs of functions are inverse of each other ?

Answer 1

One way is to solve for the invers of the first function
remembe, to solve, replace f(x) or g(x) with y, switch x and y, solve for y and replace it with f⁻¹(x)

A.
f(x)=x/2+8
y=x/2+8
x=y/2+8
x-8=y/2
2x-16=y
f⁻¹(x)=2x-16
nope, not A


B.
f(x)=3x³+16
y=3x³+16
x=3y³+16
x-16=3y³
(x-16)/3=y³
∛((x-16)/3)=y
f⁻¹(x)=∛((x-16)/3)
nope, not the same
not B

C.
f(x)=18/x-9
y=18/x-9
x=18/y-9
x+9=18/y
y(x+9)=18
y=18/(x+9)
f⁻¹(x)=18/(x+9)
correct, the answer is C



answer is C

Answer 2

Option C.

Further explanation

To find the formula for the inverse of a function, solve the equation y = f(x) for x and interchange x and y.

Let's examine the options one by one.

[Option A]

$\boxed{ \ f(x) = \frac{x}{2} + 8 \ } \ and \ \boxed{ \ g(x) = 2x - 8 \ }$

In the equation y = f(x), both sides are subtracted by 8.

$\boxed{ \ y - 8 = \frac{x}{2} \ }$

$\boxed{ \ \frac{x}{2} = y - 8 \ }$

Both sides are multiplied by 2.

$\boxed{ \ x = 2(y - 8) \ }$

$\boxed{ \ x = 2y - 16 \ }$

Let's replace $\boxed{x \ with \ f^{-1}(x)} \ and \ \boxed{y \ with \ x}$.

Thus, the inverse of $\boxed{ \ f(x) = \frac{x}{2} + 8 \ } \ is \ \boxed{\boxed{ \ f^{-1}(x) = 2x - 16 \ }}$

Therefore f(x) and g(x) are not pairs of functions that inverses of each other.

[Option B]

$\boxed{ \ f(x) = 3x^3 + 16 \ } \ and \ \boxed{ \ g(x) = \sqrt[3]{\frac{x}{3}} - 16 \ }$

In the equation y = f(x), both sides are subtracted by 16.

$\boxed{ \ y - 16 = 3x^3 \ }$

$\boxed{ \ 3x^3 = y - 16 \ }$

Both sides are divided by 3.

$\boxed{ \ x^3 = \frac{y - 16}{3} \ }$

Next, eliminate the cube on the variable by taking the cube root of both sides of the equation.

$\boxed{ \ x = \sqrt[3]{\frac{y - 16}{3}} \ }$

$\boxed{ \ x = \sqrt[3]{ \frac{y}{3} - \frac{16}{3}} \ }$

Let's replace $\boxed{x \ with \ f^{-1}(x)} \ and \ \boxed{y \ with \ x}$.

Thus, the inverse of $\boxed{ \ f(x) = 3x^3 + 16 \ } \ is \ \boxed{\boxed{ \ f^{-1}(x) = \sqrt[3]{ \frac{x}{3} - \frac{16}{3}} \ }}$

Therefore f(x) and g(x) are not pairs of functions that inverses of each other.

[Option C]

$\boxed{ \ f(x) = \frac{18}{x} - 9 \ } \ and \ \boxed{ \ g(x) = \frac{18}{x + 9} \ }$

In the equation y = f(x), both sides are added by 9.

$\boxed{ \ y + 9 = \frac{18}{x} \ }$

Both sides are multiplied by x and divided by (y + 9) or a crossing occurs between x and (y + 9).

$\boxed{ \ x = \frac{18}{y + 9} \ }$

Let's replace $\boxed{x \ with \ f^{-1}(x)} \ and \ \boxed{y \ with \ x}$.

Thus, the inverse of $\boxed{ \ \frac{18}{x} - 9 \ } \ is \ \boxed{\boxed{ \ f^{-1}(x) = \frac{18}{x + 9}} \ }}$

Therefore, f(x) and g(x) correctly attend pairs of functions that inverses of each other.

[Option D]

To be more skilled, we still eager to properly check Option D.

$\boxed{ \ f(x) = 8x^3 - 10 \ } \ and \ \boxed{ \ g(x) = \frac{x^3 + 10}{8} \ }$

In the equation y = f(x), both sides are added by 10.

$\boxed{ \ y + 10 = 8x^3 \ }$

$\boxed{ \ 8x^3 = y + 10 \ }$

Both sides are divided by 8.

$\boxed{ \ x^3 = \frac{y + 10}{8} \ }$

Next, eliminate the cube on the variable by taking the cube root of both sides of the equation.

$\boxed{ \ x = \sqrt[3]{\frac{y + 10}{8}} \ }$

As usual, let's replace $\boxed{x \ with \ f^{-1}(x)} \ and \ \boxed{y \ with \ x}$.

Thus, the inverse of $\boxed{ \ f(x) = 8x^3 - 10 \ } \ is \ \boxed{\boxed{ \ f^{-1}(x) = \sqrt[3]{\frac{x + 10}{8}} \ }}$

Therefore f(x) and g(x) are not pairs of functions that inverses of each other.

Learn more

1. The inverse of a function brainly.com/question/3225044  
2. If g(x) is the inverse of f(x), what is the value of f(g(2))? brainly.com/question/1517760
3. The composite function brainly.com/question/1691598  

Keywords: which of the following pairs of functions, the inverse of each other, replace, the equation, eliminate the cube on the variable, by taking the cube root of both sides, f(x), g(x)