# max and min value

• Mar 22nd 2009, 06:05 PM
rock candy
max and min value ( derivative)
find the max and min : y = x^3 - 8x^2 +61

max : dy/dx = 3x^2 - 16X ; dy/dx = 0

0=x(3x-16)
x = 5.33 y = -14.9

min : d2y/dx2 = 9x - 16 ; d2y/dx2 = 0

0=9x-16
x= 1.78 y = 41.3

its wrong ... im not sure what i did wrong though ( 3 significant digits btw)
• Mar 23rd 2009, 06:20 AM
stapel
Quote:

Originally Posted by rock candy
find the max and min : y = x^3 - 8x^2 +61

max : dy/dx = 3x^2 - 16X ; dy/dx = 0

0=x(3x-16)
x = 5.33 y = -14.9

How did you go from "x = 0 or 3x - 16 = 0" to "x = 5.33 and y = -14.9"?

(Would it be correct to assume, in what follows, that "X" is meant to be the same as "x"?)

Quote:

Originally Posted by rock candy
min : d2y/dx2 = 9x - 16 ; d2y/dx2 = 0

How did you go from $\, \frac{dy}{dx}\, =\, 3x^2\, -\, 16x\, "$ to $\,\frac{d^2y}{dx^2}\, =\, 9x\, -\, 16\,"$? (The derivative of 3x^2 is (2)(3x^(2-1)) = 2*3x = 6x.)

Instead, try starting with a quick graph, so you have a good idea what you're working with. Then find the critical points (in exact form, rather than decimal approximations) of the function.

Take the second derivative, find its zeroes, and then find the intervals of sign, which will prove what you already know from the graph. (Wink)