Blog #30

Well, more review I guess…let’s go over concavity and the 2nd Derivative Test.

Okay, concavity. To test concavity, you use the same steps as the first derivative test, but with the 2nd derivative. To review, here are the steps:
1. Take the 2nd derivative.
2. Set it equal to 0.
3. Solve for x.
4. Check differentiability.
5. Set up intervals.
6. Pick numbers on the intervals and plug them into the 2nd derivative.
7. If you get a positive number, it is concave up. If you get a negative number it is concave down.
8. If there is a change (like from positive to negative or vice versa), it is a point of inflection.

Ex. 1) Determine the open intervals on which the graph f(x) = -x^3 + 6x^2 – 9x – 1 is concave up or down.
f’(x) = -3x^2 + 12x – 9
f’’(x) = -6x + 12
-6x + 12 = 0
-6x = -12
x = 2
It is differentiable.
(-infinity, 2) u (2, infinity)
f’’(1) = positive #
f’’(3) = negative #
concave up: (-infinity, 2)
concave down: (2, infinity)
*Note that the question does not ask for the point of inflection, but if it did, you would say: At x = 2 there is a point of inflection.

Now, the 2nd Derivative Test. It is simply a shortcut for the 1st Derivative Test. Also, this test does not always work. If for step 7 you get a 0, the test fails. If that ever happens, you must use the 1st Derivative Test. Anyway, the steps are:
1. Take the 1st derivative.
2. Set it equal to 0.
3. Solve for x.
4. Check differentiability.
5. Take the 2nd derivative.
6. Plug in your critical points (your x-values) into the 2nd derivative.
7. If you get a positive number, it is a min. If you get a negative number, it is a max.

Ex. 2) Use the 2nd Derivative Test to find the relative extrema of f(x) = x^3 – 3x^2 + 3.
f’(x) = 3x^2 – 6x
3x^2 – 6x = 0
3x(x – 2) = 0
x = 0, 2
f’’(x) = 6x – 6
6(0) – 6 = negative #
6(2) – 6 = positive #
x = 0 is a max
x = 2 is a min