To find the relative extrema of the function
[tex]s(x)=-x^2-6x+112[/tex]
We need to find the derivative of the function, then
[tex]\frac{ds}{dx}=-2x-6[/tex]
Then we equate the derivative to zero and solve for x:
[tex]\begin{gathered} \frac{ds}{dx}=0 \\ -2x-6=0 \\ -2x=6 \\ x=-\frac{6}{2} \\ x=-3 \end{gathered}[/tex]
Therefore we have a relative extrema at x=-3 with value s=121.
Now we need to find out if this value is a maximum or minimum, to do that we need to find the derivative of the derivative.
[tex]\begin{gathered} \frac{d}{dx}(\frac{ds}{dx})=\frac{d}{dx}(-2x-6) \\ =-2 \end{gathered}[/tex]
Since the value of the second derivative is negative for every value of x, the relative extrema is a maximum.
Therefore we have a maximum at the point (-3,121).
