Find the inverse for each of the following functions. $$ s(x)=3 x+3 $$ $$ s^{-1}(x)=\square $$ $$ r(x)=\sqrt[3]{8 x+5} $$ $$ r^{-1}(x)=\square $$ $$ h(x)=9 x^{3}+4 $$ $$ h^{-1}(x)=\square $$ $$ k(x)=\frac{5}{x+10} $$ $$ k^{-1}(x)=\square $$

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1. $$ s^{-1}(x) = \frac{x - 3}{3} $$ 2. $$ r^{-1}(x) = \frac{x^3 - 5}{8} $$ 3. $$ h^{-1}(x) = \sqrt[3]{\frac{x - 4}{9}} $$ 4. $$ k^{-1}(x) = \frac{5}{x} - 10 $$ To find the inverse of a function, we need to switch the roles of x and y and solve for y. Let's go through each function one by one. 1. For $$ s(x) = 3x + 3 $$: Switch x and y: $$ x = 3y + 3 $$ Solve for y: $$ x - 3 = 3y $$ $$ y = \frac{x - 3}{3} $$ So, $$ s^{-1}(x) = \frac{x - 3}{3} $$. 2. For $$ r(x) = \sqrt[3]{8x + 5} $$: Switch x and y: $$ x = \sqrt[3]{8y + 5} $$ Cube both sides to eliminate the cube root: $$ x^3 = 8y + 5 $$ Solve for y: $$ 8y = x^3 - 5 $$ $$ y = \frac{x^3 - 5}{8} $$ So, $$ r^{-1}(x) = \frac{x^3 - 5}{8} $$. 3. For $$ h(x) = 9x^3 + 4 $$: Switch x and y: $$ x = 9y^3 + 4 $$ Solve for y: $$ 9y^3 = x - 4 $$ $$ y^3 = \frac{x - 4}{9} $$ Take the cube root of both sides: $$ y = \sqrt[3]{\frac{x - 4}{9}} $$ So, $$ h^{-1}(x) = \sqrt[3]{\frac{x - 4}{9}} $$. 4. For $$ k(x) = \frac{5}{x + 10} $$: Switch x and y: $$ x = \frac{5}{y + 10} $$ Solve for y: $$ x(y + 10) = 5 $$ $$ xy + 10x = 5 $$ $$ xy = 5 - 10x $$ $$ y = \frac{5 - 10x}{x} $$ $$ y = \frac{5}{x} - 10 $$ So, $$ k^{-1}(x) = \frac{5}{x} - 10 $$. These are the inverses of the given functions.

Find the inverse for each of the following functions. \( s(x)=3 x+3 \) \( s^{-1}(x)=\square \) \( r(x)=\sqrt[3]{8 x+5} \) \( r^{-1}(x)=\square \) \( h(x)=9 x^{3}+4 \) \( h^{-1}(x)=\square \) \( k(x)=\frac{5}{x+10} \) \( k^{-1}(x)=\square \)

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Find the inverse for each of the following functions. \( s(x)=3 x+3 \) \( s^{-1}(x)=\square \) \( r(x)=\sqrt[3]{8 x+5} \) \( r^{-1}(x)=\square \) \( h(x)=9 x^{3}+4 \) \( h^{-1}(x)=\square \) \( k(x)=\frac{5}{x+10} \) \( k^{-1}(x)=\square \)

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