PreCalculus
posted by Stacy .
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

I'll rewrite the equation.
y = f(x) = x^3 12x + 12
3 <= x >= 5 
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 
Thank you so much. This is a great help. I can see where I started to go wrong.
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