# What are the absolute extrema of  f(x)= 3x^3 − sqrt(3x^2 − 6x + 3) in [-2,9]?

Feb 9, 2017

f is differentiable and $f \uparrow$ with $x \in \left(- \infty , \infty\right)$. So, the end values |f(-2)|=27 and f(9)=2187-2sqrt6# are absolute extrema. See the illustrative Socratic graphs and explanation.

#### Explanation:

$f \left(x\right) = \sqrt{3} \left(\sqrt{3} {x}^{3} - | x - 1 |\right)$.

See the graphs of the given function and this equivalent function.

As $x \to \pm \infty , f \to \pm \infty$.

$f ' = 3 {x}^{2} - 1 > 0 , x > 1$,

$f ' = 3 , x = 1 \mathmr{and}$

$f ' = 3 {x}^{2} + 1 > 0 , x < 1$.'

In brief, $f \uparrow$with x.

Absolute minimum = lower-end value $f \left(- 2\right) | = 27$.

Absolute maximum = upper-end value $f \left(9\right) = 2187 - 2 \sqrt{6}$.

graph{3x^3-sqrt(3x^2-6x+3) [-10, 10, -5, 5]}

graph{3x^3-sqrt3|x-1| [-10, 10, -5, 5]}