For which of the following increasing functions f does (f-1)'(20) = 1/5
A f(x)= x + 5
B f(x) = x^3 + 5x + 20
C f(x) = x^5 + 5x + 14
D f(x)= e^x+ 5x + 19

Answers

Answer 1

Answer:

D f(x)= e^x+ 5x + 19

Step-by-step explanation:

D f(x)= e^x+ 5x + 19

Answer 2

We want to see for which one of the given functions, the inverse derivate evaluated in x = 20 is equal to 1/5.

The correct option is B:  f(x) = x^3 + 5x + 20

Remember that two functions f(x) and g(x) are inverses if:

f( g(x)) = g( f(x))) = x

And for the inverse derivate of a function, we have the rule:

[tex]\frac{d}{dx} [f^{-1}(f(x))] = \frac{1}{f'(x)}[/tex]

So first we need to find for what value of x each function is equal to 20, and then we need to see if the derivate of the function evaluated in that same value of x is equal to 5.

For example, for the first option we have:

A: f(x) = x + 5

We find the x-value such that the function is equal to 20.

f(15) = 15 + 5 = 20

We derivate the function.

f'(x) = 1

We evaluate the function in the x-value we got above.

f'(15) = 1

This is not the correct option, as this is not 5.

Now we need to do that for all the given options.

The only correct option will be the second one:

B: f(x) = x^3 + 5x + 20

First we find the x-value such that this is equal to 20

f(0) = 0^3 + 5*0 + 20 = 20

Then the x-value is x = 0.

Now we find the derivate of f(x).

f'(x) = 3*x^2 + 5

Now we evaluate that in the x-value we got before:

f'(0) = 3*0^2 + 5 = 5

As wanted, this is equal to 5.

Then we have:

[tex]\frac{d}{dx} [ f^{-1}( f(0))] = \frac{1}{f'(0)} \\\\\frac{d}{dx} [ f^{-1}(20)] = \frac{1}{5}[/tex]

As wanted.

If you want to learn more, you can read:

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