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

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

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- Nov 9th 2008, 10:06 AMsdaniels229find the absolute extrema of the function
f(x)=(x-3)^3 [0,4]

f(x)=x^3-12X [1,4] - Nov 9th 2008, 05:08 PMArch_Stanton
1.$\displaystyle f'(x)=[(x-3)^3]'=[x^3-9x^2+27x-27]'=3x^2-18x+27$

$\displaystyle f'(x)=0 \Longrightarrow x=3$

Notice, that the derivative has a double root from what results, that the $\displaystyle f$ function is a hyperbola with focus in $\displaystyle x=3 \in [0,4]$,

the extrema are then $\displaystyle f(0)$ and $\displaystyle f(4)$:

$\displaystyle f(0)=-27$

$\displaystyle f(4)=1$

2.$\displaystyle f'(x)=[x^3-12x]'=3x^2-12=3(x+2)(x-2)$, $\displaystyle x=2$ belongs to the interval,

all we have to do now is to check the $\displaystyle f$ function values for the variable of $\displaystyle 1$, $\displaystyle 2$ and $\displaystyle 4$ and pick the highest and the lowest one. - Nov 10th 2008, 04:15 AMsdaniels229was lost some saved me
thank you was very lost(Bow)