Let g(x)=(1)/(8)x^(3)+(1)/(2)x-(1)/(4) and let h be the inverse function of g. Notice that g(2)=(7)/(4). h^(')((7)/(4))=

Find g'(x): First, we need to find the derivative of g(x), which is g'(x). So let's differentiate g(x) = (1/8)x^3 + (1/2)x - (1/4).Calculate g'(2): g'(x) = (3/8)x^2 + 1/2.Use formula for inverse function derivative: Now, we know that g(2) = (7/4), so we need to find g'(2) to use in the formula for the derivative of the inverse function.Find g^(-1)(7/4): g'(2) = (3/8)(2)^2 + 1/2 = (3/8)(4) + 1/2 = 3 + 1/2 = 3.5 or 7/2.Plug in g'(2) into formula: The formula for the derivative of the inverse function at a point y is 1 / (g'(g^(-1)(y))). We have y = (7/4) and we need to find g^(-1)(7/4), but we already know that g(2) = (7/4), so g^(-1)(7/4) = 2.Simplify expression: Now we plug in g'(2) into the formula: h^(')((7/4)) = 1 / g'(2) = 1 / (7/2).Simplify the expression: h^(')((7/4)) = 1 / (7/2) = 2/7.

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Let
g(x)=(1)/(8)x^(3)+(1)/(2)x-(1)/(4) and let
h be the inverse function of
g. Notice that
g(2)=(7)/(4).

h^(')((7)/(4))=

Let $g(x)=\frac{1}{8} x^{3}+\frac{1}{2} x-\frac{1}{4}$ and let $h$ be the inverse function of $g$ . Notice that $g(2)=\frac{7}{4}$ . $\newline$ $h^{\prime}\left(\frac{7}{4}\right)=$

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Algebra 2

Simplify variable expressions using properties

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Q. Let

g(x)=\frac{1}{8} x^{3}+\frac{1}{2} x-\frac{1}{4}

and let

h

be the inverse function of

g

. Notice that

g(2)=\frac{7}{4}

\newline

h^{\prime}\left(\frac{7}{4}\right)=

Find $g'(x)$ : First, we need to find the derivative of $g(x)$ , which is $g'(x)$ . So let's differentiate $g(x) = \frac{1}{8}x^3 + \frac{1}{2}x - \frac{1}{4}$ .
Calculate $g'(2)$ : $g'(x) = \frac{3}{8}x^2 + \frac{1}{2}$ .
Use formula for inverse function derivative: Now, we know that $g(2) = \frac{7}{4}$ , so we need to find $g'(2)$ to use in the formula for the derivative of the inverse function.
Find $g^{-1}(\frac{7}{4}):$ $g'(2) = (\frac{3}{8})(2)^2 + \frac{1}{2} = (\frac{3}{8})(4) + \frac{1}{2} = 3 + \frac{1}{2} = 3.5$ or $\frac{7}{2}.$
Plug in $g'(2)$ into formula: The formula for the derivative of the inverse function at a point $y$ is $\frac{1}{g'(g^{-1}(y))}$ . We have $y = \frac{7}{4}$ and we need to find $g^{-1}(\frac{7}{4})$ , but we already know that $g(2) = \frac{7}{4}$ , so $g^{-1}(\frac{7}{4}) = 2$ .
Simplify expression: Now we plug in $g'(2)$ into the formula: $h^{'}\left(\frac{7}{4}\right) = \frac{1}{g'(2)} = \frac{1}{\left(\frac{7}{2}\right)}$ .
Simplify expression: Now we plug in $g'(2)$ into the formula: $h^{'}\left(\frac{7}{4}\right) = \frac{1}{g'(2)} = \frac{1}{\left(\frac{7}{2}\right)}$ .Simplify the expression: $h^{'}\left(\frac{7}{4}\right) = \frac{1}{\left(\frac{7}{2}\right)} = \frac{2}{7}$ .