Question

find the inverse for each of the following functions.\\( f(x) = 2x + 10 \\)\\( f^{-1}(x) = \\)\\( g(x) = 9x^3 - 13 \\)\\( g^{-1}(x) = \\)\\( h(x) = \frac{2}{x + 13} \\)\\( h^{-1}(x) = \\)\\( j(x) = \sqrt3{x + 9} \\)\\( j^{-1}(x) = \\)\\( \text{question help:} \\) \\( \boxed{\text{video 1}} \\) \\( \boxed{\text{video 2}} \\)\\( \text{submit question} \\)

Explanation:

For $f(x) = 2x + 10$:

Step1: Set $y=f(x)$

$y = 2x + 10$

Step2: Swap $x$ and $y$

$x = 2y + 10$

Step3: Solve for $y$

$x - 10 = 2y \implies y = \frac{x - 10}{2}$

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For $g(x) = 9x^3 - 13$:

Step1: Set $y=g(x)$

$y = 9x^3 - 13$

Step2: Swap $x$ and $y$

$x = 9y^3 - 13$

Step3: Isolate the cubic term

$x + 13 = 9y^3$

Step4: Solve for $y$

$y^3 = \frac{x + 13}{9} \implies y = \sqrt[3]{\frac{x + 13}{9}}$

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For $h(x) = \frac{2}{x + 13}$:

Step1: Set $y=h(x)$

$y = \frac{2}{x + 13}$

Step2: Swap $x$ and $y$

$x = \frac{2}{y + 13}$

Step3: Cross-multiply to isolate $y$

$x(y + 13) = 2 \implies y + 13 = \frac{2}{x} \implies y = \frac{2}{x} - 13$

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For $j(x) = \sqrt[3]{x + 9}$:

Step1: Set $y=j(x)$

$y = \sqrt[3]{x + 9}$

Step2: Swap $x$ and $y$

$x = \sqrt[3]{y + 9}$

Step3: Cube both sides

$x^3 = y + 9$

Step4: Solve for $y$

$y = x^3 - 9$

Answer:

$f^{-1}(x) = \frac{x - 10}{2}$
$g^{-1}(x) = \sqrt[3]{\frac{x + 13}{9}}$
$h^{-1}(x) = \frac{2}{x} - 13$
$j^{-1}(x) = x^3 - 9$