# Thread: find the absolute extrema of the function

1. ## find the absolute extrema of the function

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

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

2. 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.

3. ## was lost some saved me

thank you was very lost