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If
g(x)=f^(-1)(x), and
f(x)=sqrt(x-4) then solve for
g^(')(1).

$4$ . If $g(x)=f^{-1}(x)$ , and $f(x)=\sqrt{x-4}$ then solve for $g^{\prime}(1)$ .

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Find the vertex of the transformed function

Full solution

Q.

4

. If

g(x)=f^{-1}(x)

, and

f(x)=\sqrt{x-4}

then solve for

g^{\prime}(1)

.

Understand Relationship: Understand the relationship between $f(x)$ and $g(x)$ . $f(x) = \sqrt{x-4}$ and $g(x) = f^{-1}(x)$ , meaning $g(f(x)) = x$ .
Find Derivative of $f(x)$ : Find the derivative of $f(x)$ . $f'(x) = \left(\frac{1}{2}\right)\left(x-4\right)^{-\frac{1}{2}} \cdot 1 = \frac{1}{2\sqrt{x-4}}.$
Apply Inverse Function Formula: Apply the formula for the derivative of the inverse function. $g'(x) = \frac{1}{f'(g(x))}.$
Substitute $x = 1$ : Substitute $x = 1$ into $g'(x)$ . $\newline$ $g'(1) = \frac{1}{f'(g(1))}$ .
Find $g(1)$ : Find $g(1)$ using the fact that $g(f(x)) = x$ . $\newline$ Since $f(g(1)) = 1$ , $\sqrt{g(1)-4} = 1$ . $\newline$ $g(1) - 4 = 1^2$ . $\newline$ $g(1) = 5$ .
Substitute $g(1)$ into $f'(g(1))$ : Substitute $g(1)$ into $f'(g(1))$ . $\newline$ $f'(5) = \frac{1}{2\sqrt{5-4}} = \frac{1}{2}$ .
Calculate $g'(1)$ : Calculate $g'(1)$ . $\newline$ $g'(1) = \frac{1}{(\frac{1}{2})} = 2$ .

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