Question

The function h is given by h(x)=x^{5}+3 x-2 and h(1)=2 . If h^{-1} is the inverse of h , what is the value of (h^{-1})'(2) ? (A) \frac{1}{83}

Answer 1

If
`h^(-1)` is the inverse of h , option B
`(1)/(8)` is the value of
`(h^(-1))'(2)`.

To find the value of
`\left(h^(-1)\right)^(\prime)(2)`, we can use the fact that the derivative of an inverse function is the reciprocal of the derivative of the original function at the corresponding point.

Let's denote the inverse function
`h^(-1)(y)`, where y is the output value of h(x). So,
`h^(-1)(2)=1` since h(1)=2.

Now, the derivative of h(x) is given by
`h^(\prime)(x)=5 x^4+3`. We can evaluate
`h^(\prime)(1)` to find the slope of the tangent line to h(x) at x=1:

`h^(\prime)(1)=5(1)^4+3=5+3=8`

Now,
`\left(h^(-1)\right)^(\prime)(2)` is the reciprocal of
`h^(\prime)(1)`:

`\left(h^(-1)\right)^(\prime)(2)=(1)/(h^(\prime)(1))=(1)/(8)`

So, the correct answer is (B)
`(1)/(8)`

Complete Question:
The function h is given by
`h(x)=x^5+3 x-2` and
`h(1)=2`. If
`h^(-1)` is the inverse of h, what is the value of
`\left(h^(-1)\right)^(\prime)(2)` ?

a.
`(1)/(83)`

b.
`(1)/(8)`

c.
`(1)/(2)`

d. 1

e. 8

Answer 2

We can see that the derivate of the inverse is h⁻¹'(2) = 1/83

How to find h⁻¹(2)'?

The rule for the derivate of the invertse is:

h⁻¹'(x) = 1/h'(x)

Here we know that:

h(1) = 2

`h(x) = x^5 + 3x - 2`

The derivate of this is:

`h'(x) = 5x^4 + 3`

Evaluating this in x = 2 we get:

`h'(2) = 5*2^4 + 3= 5*16 + 3 = 83`

Then_

h⁻¹'(2) = 1/h'(2) = 1/83

"The function h is given by h(x)=x^{5}+3 x-2 and h(1)=2 . If h^{-1} is the inverse of h , what is the value of (h^{-1})'(2) ?

(A) \frac{1}{83}

(B) 83

(C) 1

(D) 0"