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Let
f(x)=(1)/(2)x^(3)+3x-4 and let
h be the inverse function of
f. Notice that
f(-2)=-14.

h^(')(-14)=

Let $f(x)=\frac{1}{2} x^{3}+3 x-4$ and let $h$ be the inverse function of $f$ . Notice that $f(-2)=-14$ . $\newline$ $h^{\prime}(-14)=$

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Transformations of quadratic functions

Full solution

Q. Let

f(x)=\frac{1}{2} x^{3}+3 x-4

and let

h

be the inverse function of

f

. Notice that

f(-2)=-14

.

\newline

h^{\prime}(-14)=

Find $h'(-14)$ : To find $h'(-14)$ , we need to use the formula for the derivative of the inverse function: $h'(y) = \frac{1}{f'(x)}$ where $f(x) = y$ .
Find $f'(-2)$ : Since $f(-2) = -14$ , we have $x = -2$ when $y = -14$ . Now we need to find $f'(-2)$ .
Differentiate $f(x)$ : Differentiate $f(x) = \frac{1}{2}x^3 + 3x - 4$ to get $f'(x) = \frac{3}{2}x^2 + 3$ .
Plug $x = -2$ : Now plug $x = -2$ into $f'(x)$ to find $f'(-2)$ : $f'(-2) = (\frac{3}{2})(-2)^2 + 3 = (\frac{3}{2})\cdot4 + 3 = 6 + 3 = 9$ .
Use formula for $h'(-14)$ : Use the formula for the derivative of the inverse function: $h'(-14) = \frac{1}{f'(-2)} = \frac{1}{9}$ .

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