Consider the equation below. f(x) = 2x^3 + 3x^2 - 72x Find the interval on which
ID: 3016463 • Letter: C
Question
Consider the equation below. f(x) = 2x^3 + 3x^2 - 72x Find the interval on which f is increasing. (Enter your answer in interval notation.) _________ Find the interval on which f is decreasing. (Enter your answer in interval notation.) _________ Find the local minimum and maximum values of f. local minimum _________ local maximum _______ Find the inflection point. (x, y) = (_______) Find the interval on which f concave up. (Enter your answer in Interval notation.) ________ Find the interval on which f is concave down. (Enter your answer in interval notation.)Explanation / Answer
a) interval on which f is increasing is the interval where the derivative is greater that zero
So set f'(x) > 0:
f'(x) > 0
6x^2 + 6x - 72 > 0
x^2 + x - 12 > 0
(x + 4) (x - 3) > 0
So f is increasing on the interval (-, -4) U (3, )
To find the interval where f is decreasing, set it less than zero and solve the same way
(x + 4) ( x - 3) < 0
f is decreasing on the interval (-4, 3)
b) Local minimum and local maximum:
Set the derivative equal to zero and solve for x (these will be the critical values):
6x^2 + 6x - 72 = 0
x^2 + x - 12 = 0
(x + 4)(x - 3) = 0
x = -4, 3
So the points for these values would be (-4, 108) and (3, -135)
If you wanted, you could use the second derivative test on these values, but since you already know f is increasing on the interval (-,-4) and decreasing on (-4, 3) (in other words, going up to (-4,108) and then down), you know that (-4, 108) is a local maximum.
Similarly, because the graph goes down from x = -4 to the point (3, -135) and then up, you know that (3, -135) is a local minimum.
So local maximum at (-4, 108)nd local minimum at (3, -135)
c) The inflection point will (possibly) occur at the solution for x when the second derivative is zero.
f'(x) = 6x^2 + 6x - 72
f''(x) = 12x + 6
0 = 12x + 6
12x = -6
x = -1/2
Technically, we need to find out the rest of part c before we can say for certain that the inflection point is where x = -1/2
Plug values into the second derivative to find the intervals where it is concave up or concave down (positive value = concave up, negative value = concave down):
-10: -120 + 6 < 0, so concave down on interval (-, -1/2)
10: 120 + 6 > 0, so concave up on interval (-1/2, )
Since the concavity switches at x = -1/2, it is an inflection point
So the inflection point is (-1/2, 73/2)