Question

Given the function g(x)=4x^3−24x^2+36x, find the first derivative, g'(x)
g′(x)= ???

Notice that g'(x)=0 when x=1, that is, g'(1)=0

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

Evaluate g"(1)
g''(1)=???

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

At x=1 the graph of g(x) is concave= ???

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

At x=1 there is a local= ???

Answer 1