Find all values of x for which the graph of f(x) = (x + 5)(x - 2)^2 is concave downward?

Question

A. x < -1/3
B. -5 < x < 2
C. -8/3 < x < 2
D. 0 < x
E. none of these

Answer 1

A graph is considered "concave downward" at any point where its slope (technically, the slope of the tangent line) is decreasing. To find the change in the slope, we need the 2nd derivative with respect to the variable.

f(x) = x^3 + x^2 - 16x + 20
f'(x) = 3x^2 + 2x - 16
f''(x) = 6x + 2

So we set f''(x) < 0, to show where the slope of f(x) is decreasing.

6x + 2 < 0
6x < -2
x < -1/3

So "A" should be the correct answer.

Hope that helps!

EDITED TO ADD: The person who gave "C" as the answer was incorrect, because it is *not* true that a curve is concave downward everywhere it moves from a local maximum to a local minimum.

To see why, imagine moving along the curve in the -x direction -- by the same logic, that section of the curve would have to be concave upward, because we're moving from a local min to a local max. Since it's not possible for one section of the curve to be both concave upward and concave downward, the assumption is therefore false.

Answer 2

f(x) = (x+5)( (x^2) -4x +4)
= (x^3) -4(x^2) +4x +5(x^2) -20x +20
= (x^3) +(x^2) -16x +20

f'(x) = 3(x^2) +2x -16
equalise this to zero to obtain the stationary points
3(x^2) +2x -16 = 0
(3x +8) (x -2)= 0
x= -8/3 , +2
to now which is the min. do a second differentiation
f"(x) = 6x +2
if the value of this function is positive, when substituting the x with the values obtained, then this gives the x-coordinate of the min. so substituting x=2, f"(x) = 14 so the min. has an x-coordinate of 2, while the max. has an x-coordinate of -8/3. The graph is concave downward between those two values since the curve is moving from a max. point to a min. point.

so -8/3

Answer 3

I looked at both answer given so far A and C. The problem with A is that while correct, it is not the most correct answer while C is the complete answer. And since it says find all the values of x I would have to go with C.

Answer 4

Great answer Jay H. No need for me to show the same thing.