Question

Let f(x) = x^3+3x^2-24x

(a) Find the first derivative of f(x) and identify the critical values.

f'(x)=3x^2+6x-24

Critical Values: x=2, x=-4

(b) Find the absolute extrema of f(x) on the interval [-6,0]. Classify each absolute extremum as an absolute maximum or absolute minimum.

(No idea what it is asking for!!!)

Answer 1

You already wrote f'(x) and the critical values so I'm assuming you know them already.

Are you wondering on how to solve b)? @Fakshon

Answer 2

yes

Answer 3

If you are looking for absolute extrema of f(x) on the interval [-6,0], then you're basically for the critical values of the absolute extrema that occur from [0, -6]. You already have the critical values written and the only one that falls in [-6, 0] is x = -4. So we know that this is the critical value at which an extremum occurs in the given interval. But we need to know if this is minimum or a maximum. To find that, we f''(x), and if you recall, if f''(x) > 0, then function is concave up; minimum, if f''(x) < 0, then function is concave down; maximum.

So find f''(x), and evaluate it for x = -4. That will tell if you if it's a maximum/minimum, and then just evaluate of f(x) to get the the y-coordinate for when x = -4 which gives you the critical point.

@Fakshon

Answer 4

I am getting f''(x)=6x+6 ok

Answer 5

That's correct.

Answer 6

Now evaluate it for x = 4 and check if it's a max or a min. @Fakshon