1.7 Inverse functions  (Page 8/10)

$f(x)=-3x+5$

$f(x)=\left|x-3\right|$

$f(x)=\left|x+1\right|$

$f(x)=x^2-2$

$f(2)$

$f(\mathrm{-2})$

If $f(x)=\mathrm{-2},$ then solve for $x.$

If $f(x)=1,$ then solve for $x.$

$\frac{h(2)-h(1)}{2-1}$

$\frac{h(a)-h(1)}{a-1}$

$f(x)=\frac{2}{3x+2}$

$f(x)=\frac{x-3}{x^2-4x-12}$

$f(x)=\frac{\sqrt{x-6}}{\sqrt{x-4}}$

Graph this piecewise function: $f(x)=\{\begin{array}{l}x+1x<-2\\ -2x-3x\ge -2\end{array}$

$f(x)=4x-3$

$f(x)=10x^2+x$

$f(x)=-\frac{2}{x^2}$

Find the local minimum of the function graphed in [link] .

Find the local extrema for the function graphed in [link] .

For the graph in [link] , the domain of the function is $\left[-3,3\right].$ The range is $\left[-10,10\right].$ Find the absolute minimum of the function on this interval.

Find the absolute maximum of the function graphed in [link] .

$f(x)=4-x,g(x)=-4x$

$f(x)=3x+2,g(x)=5-6x$

$f(x)=x^2+2x,g(x)=5x+1$

$f(x)=\sqrt{x+2},g(x)=\frac{1}{x}$

$f(x)=\frac{x+3}{2},g(x)=\sqrt{1-x}$

$f(x)=\frac{x+1}{x+4},g(x)=\frac{1}{x}$

$f(x)=\frac{1}{x+3},g(x)=\frac{1}{x-9}$

$f(x)=\frac{1}{x},g(x)=\sqrt{x}$

$f(x)=\frac{1}{x^2-1},g(x)=\sqrt{x+1}$

$H(x)=\sqrt{\frac{2x-1}{3x+4}}$

$H(x)=\frac{1}{{(3x^2-4)}^{-3}}$

$f(x)={(x-3)}^2$

$f(x)={(x+4)}^3$

$f(x)=\sqrt{x}+5$

$f(x)=-x^3$

$f(x)=\sqrt[3]{-x}$

$f(x)=5\sqrt{-x}-4$

$f(x)=4\left[\left|x-2\right|-6\right]$

$f(x)=-{(x+2)}^2-1$

$g(x)=f(x-1)$

$g(x)=3f(x)$

$f(x)=3x^4$

$g(x)=\sqrt{x}$

$h(x)=\frac{1}{x}+3x$