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Which pair of functions is not a pair of inverse functions?

A. f(x)= x+1/6 and g(x)= 6x-1

B. f(x)= x-4/19 and g(x)= 19x+4

C. f(x)= x5 and g(x)= 5√x

D. f(x)= x/x + 20 and g(x)= 20x/x-1
Thank you!

2 Answers

3 votes

D. f(x)= x/x + 20 and g(x)= 20x/x-1

User Porglezomp
by
8.5k points
4 votes

Answer:

Option in D is not an inverse function.

Explanation:

A. Say,
y = f(x) = (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) = (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) = (x)/(x + 20)

⇒ yx + 20y = x

⇒ x(y - 1) = - 20y


x = (- 20y)/(y - 1)

Now, if g(x) is the inverse function of f(x), then,
g(x) = (- 20x)/(x - 1)

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

User RyanJM
by
9.0k points
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