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Let
g(x)=3-2x-x^(3) and let
f be the inverse function of
g. Notice that
g(1)=0.

f^(')(0)=

Let $g(x)=3-2 x-x^{3}$ and let $f$ be the inverse function of $g$ . Notice that $g(1)=0$ . $\newline$ $f^{\prime}(0)=$

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Evaluate exponential functions

Full solution

Q. Let

g(x)=3-2 x-x^{3}

and let

f

be the inverse function of

g

. Notice that

g(1)=0

.

\newline

f^{\prime}(0)=

Find Derivative of $g(x)$ : First, find the derivative of $g(x)$ , which is $g'(x)$ . $\newline$ $g'(x) = -2 - 3x^2$
Evaluate $g'(x)$ at $x=1$ : Evaluate $g'(x)$ at $x=1$ , since $g(1)=0$ and we need the derivative at this point to find the inverse function's derivative. $\newline$ $g'(1) = -2 - 3(1)^2$ $\newline$ $g'(1) = -2 - 3$ $\newline$ $g'(1) = -5$
Calculate $f'(0)$ : The derivative of the inverse function $f$ at a point $y$ is the reciprocal of the derivative of $g$ at the point $x$ , where $g(x)=y$ . Since $g(1)=0$ , we have $f'(0) = \frac{1}{g'(1)}$ .
Calculate $f'(0)$ : The derivative of the inverse function $f$ at a point $y$ is the reciprocal of the derivative of $g$ at the point $x$ , where $g(x)=y$ . Since $g(1)=0$ , we have $f'(0) = \frac{1}{g'(1)}$ .Now, calculate $f'(0)$ using the value of $g'(1)$ . $f$ $0$ $f$ $1$

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