Answer :
Find all relative extrema.
[tex]f (x) = x^2 + 9x -4 \\f'(x)=(x^2 + 9x -4)'=2x+9 \\f'(x)=0 \\2x+9=0 \\2x=-9 \\x=- \frac{9}{2} [/tex]
Determine if relative extrema is minimum or maximum.
[tex]f''(x)=(2x+9)'=2\ \textgreater \ 0 \Rightarrow x_{min}=- \frac{9}{2} [/tex]
[tex]y_{min}=(- \frac{9}{2} )^2+9(- \frac{9}{2})-4= \frac{81}{4}- \frac{81}{2} -4=\frac{81}{4}- \frac{162}{4} - \frac{16}{4} =- \frac{97}{16} \\ \\(x,y)=(- \frac{9}{2},- \frac{97}{16} )[/tex]
[tex]f (x) = x^2 + 9x -4 \\f'(x)=(x^2 + 9x -4)'=2x+9 \\f'(x)=0 \\2x+9=0 \\2x=-9 \\x=- \frac{9}{2} [/tex]
Determine if relative extrema is minimum or maximum.
[tex]f''(x)=(2x+9)'=2\ \textgreater \ 0 \Rightarrow x_{min}=- \frac{9}{2} [/tex]
[tex]y_{min}=(- \frac{9}{2} )^2+9(- \frac{9}{2})-4= \frac{81}{4}- \frac{81}{2} -4=\frac{81}{4}- \frac{162}{4} - \frac{16}{4} =- \frac{97}{16} \\ \\(x,y)=(- \frac{9}{2},- \frac{97}{16} )[/tex]