Page 53 - Calc for the Clueless

It is almost always easier to find the y value from the factored equation of y.

We will use both tests to test points. However, when you do the problem, once you get an answer for one point,
go on to the next point.

We first substitute x = 0,3 into f"(x) = 12x(x - 2). f"(0) = 0, so this test fails, f"(3) is positive, so (3,-27) is a
minimum.

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TEST 2 f'(x) = 4x (x - 3). f'(0 ) is negative and f'(0 ) is negative. (0,0) is neither a max nor a min. f'(3 ) is
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negative and f'(3 ) is positive. Again, (3,-27) is a minimum.
Remember, always substitute into the factored form. When testing, do not evaluate. You are only interested in
the sign of the answer.
Possible inflection points f"(x) = 0 = 12x(x - 2). x = 0, x = 2. Substituting back into the original equation
(factored) for y, we get the points (0,0) and (2,-16).

Let us use both tests.

TEST 1 f'"(0) and f'"(2) are both not 0. Therefore (0,0) and (2,-16) are inflection points.

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TEST 2 f"(0 ) is positive and f"(0 ) is negative. Different signs. (0,0) is an inflection point, f"(2 ) is negative and
f"(2 ) is positive. Different signs. (2,-16) is an inflection point.
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What happens to the ends? For large x,y = (approximately) = x . f(100) is positive and f(-100) is positive. Both
ends go to plus infinity.
Now connect the dots.

Example 25—