Q: Consider the function: f(x) = (1/3) x3 – x2 – 15x + 1
Find / tell me about:
(a) critical points
(b) the local minimum / maximums [relative extrema]
(c) points of inflection
A: OK. Get out your pen and paper and be writing with me as I calculate and explain. This is the best way for you to learn.
(a) The critical points are where the derivative equals 0 or is undefined. So, take the derivative to start:
f(x) = (1/3) x3 – x2 – 15x + 1
f ‘ (x) = x2 – 2x- 15 <– you should get this!
Now, we need to find where the derivative = 0 (or is undefined — since the derivative is a quadratic, it is never undefined, so we don’t have to worry about that part).
0 = x2 – 2x- 15
Solve by factoring:
0 = (x – 5)(x + 3)
So, the critical points are when x = 5, -3.
(b) To determine if the critical points are local maximum or minimums, we need to do the second derivative test.
Recap on second derivative test: Plug a critical point into the second derivative.
*If the end result is positive, this means the shape is concave up (smiley face) which means we have a minimum.
*If the end result is negative, this means the shape is concave down (frowny face) which means we have a maximum.
So, take the second derivative:
f(x) = (1/3) x3 – x2 – 15x + 1 [original function]
f ‘ (x) = x2 – 2x- 15 [1st derivative]
f ” (x) = 2x – 2 [2nd derivative]
Now, we have critical points of x = 5 and x = -3. Deal with them one at a time in the second derivative to perform the second derivative test.
f ” (5) = 2(5) – 2 = 10 – 2 = 8 <– this is positive, which means when x = 5 we have a local minimum!
f ” (-3) = 2(-3) – 2 = -6 – 2 = -8 <– this is negative, which means when x = -3 we have a local maximum!
If you wanted to find the (x, y) point of the actual max and min, you would plug the x-values into the original equations to find the corresponding y values:
f(5) = (1/3) (5)3 – (5)2 – 15(5) + 1 = -172/3
So, f(x) has a local minimum at the point (5, -172/3)
f(-3) = (1/3) (-3)3 – (-3)2 – 15(-3) + 1 = 28
So, f(x) has a local maximum at the point (-3, 28)
(c) The points of inflection are where the concavity of the function changes from concave up to concave down (and vice versa). Our POI candidates can be found by setting the second derivative = 0 and finding the x candidates:
f ” (x) = 2x – 2
0 = 2x – 2 and solve for x:
x = 1
So, x = 1 is a candidate for a point of inflection (POI) — it is not a guaranteed POI though.
We need to interval test into the second derivative to see what is happening (generally, you would need to test in between every x value. We only have 1 x value, so we just need to interval test on either side of it):
Plot your x values on a number line. In our case, the only value is x = 1. Pick a number on the left side of 1. Let’s pick an easy number, 0:
Plug 0 into the second derivative:
f ” (0) = 2(0) – 2 = -2 <– since this is negative: this interval, to the left of x = 1, is concave down
Now, pick a number to the right of x = 1. Let’s pick 2:
f ” (2) = 2(2) – 2 = 2 <– since this is positive: this interval, to the right of x = 1, is concave up
Therefore, when x = 1 we do have a point of inflection (since we changed from concave down to concave up).
If you want the (x, y) point, plug the x value into the original equation to find y:
f(1) = (1/3) (1)3 – (1)2 – 15(1) + 1 = -44/3
There is a point of inflection at (1, -44/3).
(d) Now to analyze concavity, which we have basically already done in part (c)!
To the left of x = 1, we determined the graph to be concave down.
To the right of x = 1, we determined the graph to be concave up.
Concave down on the interval (-infinity, 1)
Concave up on the interval (1, infinity)
We have done it. That was a lot to do. Enjoy.