Posted by **Stacy** on Saturday, November 19, 2011 at 2:28pm.

I'm having problems solving this and showing the steps:

Find the absolute maximum and absolute minimum, if either exists, for

y = f(x)= x^3 - 12x + 12 -3 less equal to x greater than equal to 5

- Pre-Calculus -
**Stacy**, Saturday, November 19, 2011 at 2:59pm
I'll rewrite the equation.

y = f(x) = x^3 -12x + 12

-3 <= x >= 5

- Pre-Calculus -
**Damon**, Saturday, November 19, 2011 at 3:33pm
dy/dx = 3x^2 -12

function is horizontal when dy/dx = 0

that is when

x^2 = 4

x = +/- 2

when x = -2

f(-2) = -8 + 24 + 12 = 28

and

d^2y/dx^2 = 6 x = -12 so a maximum

when x = 2

f(2) = -4

and

d^2y/dx^2 = 6x = +12 so a minimum

now check the end points

when x = -3

f(-3) = -27+36+12 = 21

when x=+5

f(5) = 125 - 60 +12 = 77

so the maximum is 77 at x = 5

and the minimum is -4 at x = 2

- Pre-Calculus -
**Stacy**, Saturday, November 19, 2011 at 3:55pm
Thank you so much. This is a great help. I can see where I started to go wrong.

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