find the absolute extrema of the function

**sdaniels229** · November 9th 2008, 10:06 AM

f(x)=(x-3)^3 [0,4]

f(x)=x^3-12X [1,4]

**Arch_Stanton** · November 9th 2008, 05:08 PM

1. $f'(x)=[(x-3)^3]'=[x^3-9x^2+27x-27]'=3x^2-18x+27$
$f'(x)=0 \Longrightarrow x=3$
Notice, that the derivative has a double root from what results, that the $f$ function is a hyperbola with focus in $x=3 \in [0,4]$ ,
the extrema are then $f(0)$ and $f(4)$ :
$f(0)=-27$
$f(4)=1$

2. $f'(x)=[x^3-12x]'=3x^2-12=3(x+2)(x-2)$ , $x=2$ belongs to the interval,
all we have to do now is to check the $f$ function values for the variable of $1$ , $2$ and $4$ and pick the highest and the lowest one.

**sdaniels229** · November 10th 2008, 04:15 AM

thank you was very lost

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