max and min values

Re: max and min values

Well, I'm not going to be doing much curve sketching here, but I can do the rest:

y = 4x^3 + 9x^2 - 12x -18
y' = 12x^2 + 18x -12
At extrema, y' = 0, or 12x^2 + 18x -12 = 0.
Dividing through by 6 to simplify, 2x^2 + 3x -2 = 0.
Now using the quadratic formula:

y =
-3 +/- SQRT(9 + 16)
-------------------------
2 * 2
= -3 +/- 5
----------
4

= -2, 1/2

Let's have a peek at the second derivative to determine concavity:

y' = 12x^2 + 18x -12
y'' = 24x +18
This is negative for x < -3/4 and positive for x > -3/4

Therefore, it is negative at x = -2, the first root, and positive for x = 1/2, the second root. Therefore, the first root is a relative maximum, and the second a relative minimum. The second derivative is always zero at an inflection point, since it changes sign there, so we know we don't have one of those. Since the highest power of the original equation is 3, an odd number, we know that there are only relative extrema, since the curve approaches negative infinity on the left and positive infinity on the right. Let us substitute the roots into y:

y = 4x^3 + 9x^2 - 12x -18

For x = -2, we have: -32 + 36 + 24 - 18 = 10.
For x = 1/2, we have: 1/2 + 9/4 - 6 - 18 = 11/4 - 24 = 11/4 - 96/4
= -85/4 = -21 1/4.

If this were my own homework, I really would graph it, taking into account what I have learned above, since having a visual of the thing tells a lot.

I'm going to correct myself about one thing. It's been about 30 years since I studied this, so my recollection is imperfect. The first derivative does not have to be zero for a point to be called a point of inflection, as long as the second derivative is zero and changes sign at that point, so -3/4 is one.