Given the function $\displaystyle {g{{\left({x}\right)}}}={8}{x}^{{3}}-{36}{x}^{{2}}-{96}{x}$, find the first derivative, $\displaystyle {g}'{\left({x}\right)}$.
$\displaystyle {g}'{\left({x}\right)}=$Preview Question 6 Part 1 of 5  

Notice that $\displaystyle {g}'{\left({x}\right)}={0}$ when $\displaystyle {x}={4}$, that is, $\displaystyle {g}'{\left({4}\right)}={0}$.

Now, we want to know whether there is a local minimum or local maximum at $\displaystyle {x}={4}$, so we will use the second derivative test.
Find the second derivative, $\displaystyle {g}{''}{\left({x}\right)}$.
$\displaystyle {g}{''}{\left({x}\right)}=$Preview Question 6 Part 2 of 5  

Evaluate $\displaystyle {g}{''}{\left({4}\right)}$.
$\displaystyle {g}{''}{\left({4}\right)}=$

Based on the sign of this number, does this mean the graph of $\displaystyle {g{{\left({x}\right)}}}$ is concave up or concave down at $\displaystyle {x}={4}$?
At $\displaystyle {x}={4}$ the graph of $\displaystyle {g{{\left({x}\right)}}}$ is Select an answer Concave Up Concave Down

Based on the concavity of $\displaystyle {g{{\left({x}\right)}}}$ at $\displaystyle {x}={4}$, does this mean that there is a local minimum or local maximum at $\displaystyle {x}={4}$?
At $\displaystyle {x}={4}$ there is a local Select an answer Minimum Maximum