Question

Given the function g(x)=4x3−24x2−60xg(x)=4x3-24x2-60x, find the first derivative, g'(x)g′(x).
g'(x)=g′(x)=   

Notice that g'(x)=0g′(x)=0 when x=−1x=-1, that is, g'(−1)=0g′(-1)=0.

Now, we want to know whether there is a local minimum or local maximum at x=−1x=-1, so we will use the second derivative test.
Find the second derivative, g''(x)g′′(x).
g''(x)=g′′(x)=   

Evaluate g''(−1)g′′(-1).
g''(−1)=g′′(-1)=

Based on the sign of this number, does this mean the graph of g(x)g(x) is concave up or concave down at x=−1x=-1?
[Answer either up or down -- watch your spelling!!]
At x=−1x=-1 the graph of g(x)g(x) is concave

Based on the concavity of g(x)g(x) at x=−1x=-1, does this mean that there is a local minimum or local maximumat x=−1x=-1?
[Answer either minimum or maximum -- watch your spelling!!]
At x=−1x=-1 there is a local

Answer 1