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Find all relative extrema. Use the Second Derivative Test where applicable. (If an answer does not exist, enter DNE.)

f (x) = x2 + 9x − 4 relative minimum

(x, y) =

relative maximum (x, y) =


please help me

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Verified Answer :

Find all relative extrema.

\(f (x) = x^2 + 9x -4 \\f'(x)=(x^2 + 9x -4)'=2x+9 \\f'(x)=0 \\2x+9=0 \\2x=-9 \\x=- \frac{9}{2}\)

Determine if relative extrema is minimum or maximum.

\(f''(x)=(2x+9)'=2\ \textgreater \ 0 \Rightarrow x_{min}=- \frac{9}{2}\)
\(y_{min}=(- \frac{9}{2} )^2+9(- \frac{9}{2})-4= \frac{81}{4}- \frac{81}{2} -4=\frac{81}{4}- \frac{162}{4} - \frac{16}{4} =- \frac{97}{16} \\ \\(x,y)=(- \frac{9}{2},- \frac{97}{16} )\)

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