3.3

1

Simply flip the coords so $(a,b)\to (b,a)$ then graph

2

to graph the inverse, you would flip each pair of x,y coords so for the inverse, it would be $(x,y)\to (y,x)$

3

$$\begin{array}{rl}& f(x)=2x^2-1\\ & \\ & x=2y^2-1\\ & x+1=2y^2\\ & \frac{x+1}{2}=y^2\\ & \sqrt{\frac{x+1}{2}}=y\\ & y=\sqrt{\frac{x+1}{2}}\end{array}$$

4

given

$$\begin{array}{rl}& f(x)=7-2(x-1{)}^2,x{\ge }_1\end{array}$$

determine

$f(3)$

$$\begin{array}{rl}& 7-2(3-1{)}^2\\ & 7-2(2{)}^2\\ & 7-2(4)\\ & 7-8\\ & f(3)=-1\end{array}$$

$f^{-1}(x)$

$$\begin{array}{rl}& f(x)=7-2(x-1{)}^2,x{\ge }_1\\ & x-7=-2(y-1{)}^2\\ & x-7=-2(y-1{)}^2\\ & \frac{x-7}{-2}=(y-1{)}^2\\ & \pm \sqrt{-\frac{x-7}{2}}=\sqrt{(y-1{)}^2}\\ & \pm \sqrt{-\frac{x-7}{2}}=y-1\\ & \pm \sqrt{-\frac{x-7}{2}}+1=y\\ & \pm \sqrt{\frac{7-x}{2}}+1=y\\ & y=\pm \sqrt{\frac{7-x}{2}}+1\end{array}$$

$f^{-1}(5)$

$$\begin{array}{rl}& \sqrt{\frac{7-x}{2}}+1=y\\ & \sqrt{\frac{7-5}{2}}+1=y\\ & \sqrt{\frac{2}{2}}+1=y\\ & \sqrt{1}+1=y\\ & 1+1=y\\ & y=\pm 2\end{array}$$

$f^{-1}(2a+7)$

$$\begin{array}{rl}& \sqrt{\frac{7-x}{2}}+1=y\\ & \sqrt{\frac{7-2a+7}{2}}+1=y\\ & \sqrt{\frac{2a}{2}}+1=y\\ & \sqrt{a}+1=y\end{array}$$

5

1. I would just plug in values to the formula and plot that
2. swap x,y to y,x