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Let
h(x)=(1)/(4)x^(3)+2x-1 and let
g be the inverse function of
h. Notice that
h(2)=5.

g^(')(5)=

Let $h(x)=\frac{1}{4} x^{3}+2 x-1$ and let $g$ be the inverse function of $h$ . Notice that $h(2)=5$ . $\newline$ $g^{\prime}(5)=$

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Transformations of quadratic functions

Full solution

Q. Let

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

and let

g

be the inverse function of

h

. Notice that

h(2)=5

.

\newline

g^{\prime}(5)=

Use Inverse Function Formula: To find $g'(5)$ , we need to use the formula for the derivative of the inverse function: $g'(h(x)) = \frac{1}{h'(g(x))}.$
Find $h'(x)$ : First, we need to find $h'(x)$ , which is the derivative of $h(x)$ with respect to $x$ . $\newline$ $h'(x) = \frac{1}{4} \cdot 3x^2 + 2$ .
Calculate $h'(2)$ : Now we plug in $x = 2$ into $h'(x)$ to find $h'(2)$ . $\newline$ $h'(2) = (\frac{1}{4}) \cdot 3 \cdot 2^2 + 2 = (\frac{1}{4}) \cdot 3 \cdot 4 + 2 = 3 + 2 = 5$ .
Apply Inverse Function Formula: Now we use the formula for the derivative of the inverse function. $g'(5) = \frac{1}{h'(2)} = \frac{1}{5}$ .

More problems from Transformations of quadratic functions

Question

h

is defined over the real numbers. This table gives a few values of

h

\newline

\begin{tabular}{lllll}

\newline

x

-6

1

-6

01

-6

001

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9

\newline

h(x)

-0

25

-0

74

-0

98

-1

0

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What is a reasonable estimate for

\lim _{x \rightarrow-6} h(x)

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1

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-6

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-2

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-1

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(D) The limit doesn't exist

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