Inverse Functions

Sol 1 Let $y=5x-8$. Then $y+8=5x$, so $x=1/5(y+8)$ and therefore $f^{-1}(y)=1/5(y+8)$.

Sol 2 Let $y=2x^3-5$. Then $y+5=2x^3$, so $x^3=\frac{y+5}{2}$ and $x=\sqrt[3]{\frac{y+5}{2}}$. Thus $f^{-1}(y)=\sqrt[3]{\frac{y+5}{2}}$.

Sol 3 Let $y=\sqrt{4x-9}$, so $y^2=4x-9$ and $y^2+9=4x$. Then $x=1/4(y^2+9)$, so $f^{-1}(y)=1/4(y^2+9)$. The domain of $f^{-1}$ is the same as the range of $f$, so it is the interval $[0,\infty)$

since the graph of f is the top half of the parabola $x=1/4(y^2+9)$.

Sol 4 Let $y=\frac{2x-5}{x+4}$. Then $(x+4)y=2x-5 \Rightarrow$ $xy+4y=2x-5 \Rightarrow 4y+5=2x-yx \Rightarrow 4y+5=(2-y)x \Rightarrow$ $(2-y)x=4y+5 \Rightarrow x=\frac{4y+5}{2-y}$, so $f^{-1}(y)= \frac{4y+5}{2-y}$

Sol 5 Setting $f(x)=5$, for example, and then solving for $x$ gives $f(1)=5=f(4)$; so $f$ is not 1-1 and therefore $f$ does not have an inverse.

Sol 6 Let $y-x^2-4x+9$; then $y-9=x^2-4x$, so adding 4 to both sides gives $y-5=x^2-4x+4$ and therefore $y-5=(x-2)^2$. Taking square roots of both sides, using the fact that $x\ge 2$ so $x-2\ge 0$, gives $x-2=\sqrt{y-5}$ and so $x=\sqrt{y-5}+2$. Therefore $f^{-1}(y)=\sqrt{y-5}+2$. The domain of $f^{-1}$ is the range of $f$, which is the interval $[5,\infty)$ since $f(x)=(x-2)^2+5$ for $x\ge 2$.

Sol 7 To find $f^{-1}(1/2)$, we must solve the equation $f(x)=1/2$ for $x$:

$\frac{x}{x^2-4}=1/2 \Rightarrow 2x=x^2-4 \Rightarrow 4=x^2-2x \Rightarrow$ $x^2-2x+1=5 \Rightarrow (x-1)^2=5 \Rightarrow x-1=\pm \sqrt{5}$ $\Rightarrow x=1\pm \sqrt{5} \Rightarrow x=1-\sqrt{5}$ since $-2<x<2$. Therefore $f^{-1}(1/2)=1-\sqrt{5}$.

Sol 8 Let $y=\frac{8}{\sqrt{x}+2}$.

Then $y(\sqrt{x}+2)=8 \Rightarrow y\sqrt{x}+2y=8 \Rightarrow y\sqrt{x}=8-2y$

$\Rightarrow\sqrt{x}=\frac{8-2y}{y}=\frac{8}{y}-2\Rightarrow x=(\frac{8}{y}-2)^2$,

so $f^{-1}(y)=(\frac{8}{y}-2)^2$.

The domain of $f^{-1}$ is the same as the range of $f$. Since $\sqrt{x}+2\ge 2$ for $x\ge 0$, $0<f(x)\le 4$ for $x\ge 0$. Furthermore, if $0<y\le 4$, then $\frac{8}{y}-2\ge 0$ and therefore $f((\frac{8}{y}-2)^2)=\frac{8}{(8/y-2)+2}=\frac{8}{8/y}=y$; so the domain of $f^{-1}$ is $(0,4]$,the range of $f$.

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