Calculus How do I justify this?

On the interval [-6,6], f(x) is continuous and differentiable. If f ‘(x) = (x2 – 4)(x + 1)2, briefly justify the following conclusion:
“f ‘(x) has an x-intercept at x = -1 but x = -1 is not a relative extrema on the graph of f(x) because…”

Question

Calculus How do I justify this?

On the interval [-6,6], f(x) is continuous and differentiable. If f ‘(x) = (x2 – 4)(x + 1)2, briefly justify the following conclusion: 
“f ‘(x) has an x-intercept at x = -1 but x = -1 is not a relative extrema on the graph of f(x) because…”

amoodarya · Answer

attachment

anonymous · Answer

but shouldn't -1 be a minimum??

amoodarya · Answer

you say between -1 and 1
so x=-1 is not min

anonymous · Answer

so between -1 and 1 is the minimum and this is why x=-1 is not min . correct? @amoodarya

amoodarya · Answer

yes it is not min

anonymous · Answer

thank you

phi · Answer

Notice that
$$  f ‘(x) = (x^2 – 4)(x + 1)^2$$
has a squared (x+1)... which changes your signs...

phi · Answer

but x = -1 is not a relative extrema on the graph of f(x) because…

it is neither a max nor a min, because those occur at the boundaries of your interval

anonymous · Answer

great. thanks