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

1 Answer
Feb 9, 2017

Answer:

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]}