You mean and , of course.

Just checking at one point is not in general sufficient. Here it is- if you knew that, you should have said how you knew. Also, you haven't answered the question! You haven't said what kind of point that is.So the Stationary points are and

b) Classify the LH Stationary point with 1st Derivative Test

To classify we take X = -5 and choose the test point XL = -6 then we have:

f'= 3(x-1)(x+5). If x is close to -5, then x-1 is close to -5-1= -4 and so is negative. For x< -5, x+ 5< 0 so f'= 3(x-1)(x+5) is the product of one positive and two negative numbers- f' is positive so f is increasing coming up to -5. For x> -5, x+5> 0 so f'= 3(x-1)(x+5) is the product of two positive and one negative number. f'< 0 so f is decreasing past -5. Since f was increasing and then started decreasing, x= -5 is amaximum.

No, no, no! For the second derivative test, you evaluate f"c) Classify the RH Stationary point with 2nd Derivative Test

So,

The stationary point XR = 1, so the test point will be XR = 2

atthe stationary point! f"(1)= 6(1+2)= 3> 0.

Yes, but you were lucky!So f has a local minimum at X = 1

Why are you taking points exactly "1" away from the stationary point? There is nothing special about integers!d) Find the coordinates of each stationary point

so,

Hence the graph passes through the point (-5,76) & (1,-32)

Thanks for your help

A