Question

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

Answer 1

f is differentiable and `f uarr` with `x in (-oo, oo)`. So, the end values |f(-2)|=27 and f(9)=2187-2sqrt6# are absolute extrema. See the illustrative Socratic graphs and explanation.

Explanation:

`f(x)=sqrt3(sqrt3x^3-|x-1|)`.

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

As `x to +-oo, f to +-oo`.

`f'=3x^2-1>0, x >1`,

`f'=3, x =1 and`

`f'=3x^2+1>0, x <1`.'

In brief, `f uarr`with x.

Absolute minimum = lower-end value `f(-2)|=27`.

Absolute maximum = upper-end value `f(9) =2187-2sqrt6`.

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

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