Question

Please help!

Answer 1

Answer:

Option in D is not an inverse function.

Step-by-step explanation:

A. Say, $y = f(x) = \frac{x + 1}{6}$

⇒ 6y = x + 1

⇒ x = 6y - 1

Here, g(x) = 6x - 1, if g(x) is the inverse function of f(x).

B. Let us assume, $y = f(x) = \frac{x - 4}{19}$

⇒ 19y = x - 4

⇒ x = 19x + 4

If g(x) is the inverse function of f(x), then, g(x) = 19x + 4

C. Let $y = f(x) = x^{5}$

⇒ $x = \sqrt[5]{y}$

Here, $g(x) = \sqrt[5]{x}$, if g(x) is the inverse function of f(x).

D. Let $y = f(x) = \frac{x}{x + 20}$

⇒ yx + 20y = x

⇒ x(y - 1) = - 20y

⇒ $x = \frac{- 20y}{y - 1}$

Now, if g(x) is the inverse function of f(x), then, $g(x) = \frac{- 20x}{x - 1}$

Therefore, option in D is not an inverse function. (Answer)