Find the inverse of a function - High school Precalculus Exercises

Find the inverse g of the following functions:

For x≥0, f\left(x\right)=3x^{2}-4

g is defined for x≥-4, and g\left(x\right)=\sqrt{\dfrac{x+4}{3}}

g is defined for x≥0, and g\left(x\right)=\sqrt{\dfrac{x-4}{3}}.

g is defined for x\leqslant0, and g\left(x\right)=-\sqrt{\dfrac{x+4}{3}}

g is defined for x\leqslant-4, and g\left(x\right)=\dfrac{1}{3}\sqrt{x+4}

For x≥3, f\left(x\right)=2\sqrt{x-3}+7

g is defined for x≥3, and g\left(x\right)=\left( \dfrac{x-7}{2} \right)^{2}+3

g is defined for x≥3, and g\left(x\right)=\left( \dfrac{x+7}{2} \right)^{2}-3.

g is defined for x≥7, and g\left(x\right)=\left( \dfrac{x+7}{2} \right)^{2}-3

g is defined for x≥7, and g\left(x\right)=\left( \dfrac{x-7}{2} \right)^{2}+3

For x \gt 8, f\left(x\right)=3\log_5\left(x-8\right)+1

g is defined for x \gt 1, and g\left(x\right)=5^{\frac{x-1}{3}}-8

g is defined for x\in\mathbb{R}, and g\left(x\right)=5^{\frac{x-1}{3}}+8

g is defined for x \gt 8, and g\left(x\right)=5^{\frac{x}{3}-1}+8

g is defined for x \gt 1, and g\left(x\right)=\dfrac{5^{x-1}}{3}+8

For x\in \mathbb{R}, f\left(x\right)=4e^{3x}-9

g is defined for x\geqslant-9, and g\left(x\right)=\dfrac{1}{3}\ln\left(\dfrac{x+9}{4}\right)

g is defined for x\geqslant-9, and g\left(x\right)=\dfrac{1}{3}e^{\frac{x+9}{4}}

g is defined for x\geqslant9, and g\left(x\right)=3\ln\left(\dfrac{x-9}{4}\right)

g is defined for x\in\mathbb{R}, and g\left(x\right)=\ln\left(\dfrac{x+9}{4}\right)^3

For x\neq-3, f\left(x\right)=\dfrac{2x-4}{x+3}

g is defined for x\neq-3, and g\left(x\right)=\dfrac{4x-2}{x+3}

g is defined for x\neq2, and g\left(x\right)=\dfrac{x+3}{2x-4}

g is defined for x\neq-4, and g\left(x\right)=\dfrac{2x-3}{x+4}

g is defined for x\neq2, and g\left(x\right)=\dfrac{3x+4}{2-x}

For x\in\mathbb{R}, f\left(x\right)=5x^{3}+9

g is defined for x\in\mathbb{R}, and g\left(x\right)=\sqrt[3]{\frac{x-9}{5}}

g is defined for x\geqslant9, and g\left(x\right)=\dfrac{1}{5}\sqrt[3]{x-9}

g is defined for x\geqslant-9, and g\left(x\right)=\sqrt[3]{\frac{x+9}{5}}

For x\in\mathbb{R}, f\left(x\right)=4\sqrt[5]{2x+3}-7

g is defined for x\geqslant-7, and g\left(y\right)=\dfrac{1}{2}\left[ \left( \dfrac{y-7}{4} \right)^{5}+3 \right]

g is defined for x\in\mathbb{R}, and g\left(y\right)=\dfrac{1}{2}\left[ \left( \dfrac{y-7}{4} \right)^{5}+3 \right]

g is defined for x\in\mathbb{R}, and g\left(y\right)=\dfrac{1}{2}\left[ \left( \dfrac{y+7}{4} \right)^{5}-3 \right]

g is defined for x\geqslant-\dfrac{3}{2}, and g\left(y\right)=\dfrac{1}{4\sqrt[5]{2x+3}-7}