Given the function $g(x)=6 x^{3}+27 x^{2}-180 x$, find the first derivative, $g^{\prime}(x)$. $g^{\prime}(x)=$ Notice that $g^{\prime}(x)=0$ when $x=-5$, that is, $g^{\prime}(-5)=0$. Now, we want to know whether there is a local minimum or local maximum at $x=-5$, so we will use the second derivative test. Find the second derivative, $g^{\prime \prime}(x)$. \[ g^{\prime \prime}(x)= \] Evaluate $g^{\prime \prime}(-5)$ \[ g^{\prime \prime}(-5)= \] Based on the sign of this number, does this mean the graph of $g(x)$ is concave up or concave down at $x=-5$ ? [Answer either up or down -- watch your spelling!!] At $x=-5$ the graph of $g(x)$ is concave Based on the concavity of $g(x)$ at $x=-5$, does this mean that there is a local minimum or local maximum at $x=-5$ ? [Answer either minimum or maximum -- watch your spelling!!] At $x=-5$ there is a local

Step 1 ：Given the function \(g(x)=6 x^{3}+27 x^{2}-180 x\), find the first derivative, \(g^{\prime}(x)\).

Step 2 ：By applying the power rule of differentiation, the first derivative of the function \(g(x)\) is \(g^{\prime}(x) = 18x^2 + 54x - 180\).

Step 3 ：Setting the first derivative equal to zero and solving for \(x\), we find that \(x=-5\) is a critical point.

Step 4 ：Find the second derivative, \(g^{\prime \prime}(x)\), by differentiating the first derivative. The second derivative of the function is \(g^{\prime \prime}(x) = 36x + 54\).

Step 5 ：Evaluate the second derivative at \(x = -5\), which gives a value of \(g^{\prime \prime}(-5) = -126\).

Step 6 ：Since the second derivative at \(x = -5\) is negative, the function is concave down at this point. Therefore, at \(x=-5\) the graph of \(g(x)\) is concave \(\boxed{down}\).

Step 7 ：Because the function is concave down at \(x = -5\), there is a local maximum at this point. Therefore, at \(x=-5\) there is a local \(\boxed{maximum}\).