## Precalculus (6th Edition) Blitzer

(a) f has a relative maximum at c if there is some interval $\left( r,s \right)$ containing c for which $f\left( c \right)\ge f\left( x \right)$ for all $x$ between r and s for which $f\left( x \right)$ is defined. Therefore, at the point $x=0$, the graph gives the relative maximum of $-2$. Hence, the relative maxima occurs at $x=0$, and graph gives the relative maximum of $f\left( 0 \right)=-2$. (b) In case of the interval $\left( r,s \right)$ containing c for which $f\left( c \right)\le f\left( x \right)$ for all $x$ between r and s for which $f\left( x \right)$ is defined, f will have a relative minimum at c. Therefore, at points $-2$ and $3,$ the graph gives the relative minima of $-3,-5,$ respectively. Hence, the relative maxima occurs at $x=-2,3;f\left( -2 \right)=-3,f\left( 3 \right)=-5$ and at points $-2$ and $3,$ the graph gives the relative minima of $-3,-5,$ respectively.