# Find the abs max and min of the function SHOW ALL WORK

label Calculus
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Jul 18th, 2015

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$f(x)=x^3+3x^2+3\\$

Differentiate f(x) w.r.t. x

$\\ f'(x)=3x^2+3\times2x\\ \\ f'(x)=3x^2+6x\\$

Put f'(x) = 0

$\\ 3x^2+6x=0\\ \\ 3x(x+2)=0\\ \\ 3x=0\hspace{5}or\hspace{5}x+2=0\\ \\ x=0\hspace{5}or\hspace{5}x=-2$

So, we have two critical points x = 0 and x = -2.

Since x = -2 does not lie in the interval [-1, 2], it can be ignored.

Now to find the absolute maximum and minimum in [-1, 2] we have to evaluate the value of the function at the critical point and at the end points.

$\\ f(-1)=(-1)^3+3(-1)^2+6=-1+3+6=8\\ \\ f(0)=0^3+3(0)^2+6=0+0+6=6\\ \\ f(2)=2^3+3(2)^2+6=8+12+6=26\\$

So, absolute maximum of f(x) is 26 which occurs at x = 2 (an endpoint)

and absolute minimum of f(x) is 6 which occurs at x = 0 (a critical point)

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Jul 18th, 2015

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Jul 18th, 2015
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Jul 18th, 2015
Nov 22nd, 2017
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