Question 1155974
{{{f(x)= 2x^3+3x^2-120x+2}}}

[{{{-5}}},{{{7}}}]


a. Locate the critical points of the given function.


find first derivative


{{{f}}}'{{{ (x) =6x^2+6x-120}}}

set {{{f}}}'{{{ (x) =0}}}


{{{6x^2+6x-120=0}}}

{{{6(x^2+x-20)=0}}}

will be zero if

{{{x^2+x-20=0}}}.....factor

{{{x^2-4x+5x-20=0}}}

{{{(x^2-4x)+(5x-20)=0}}}

{{{x(x-4)+5(x-4)=0}}}

{{{(x - 4) (x + 5) = 0}}}


=> extreme points are at


{{{x=4}}}  and  {{{x=-5}}}


use second derivative test to determine where is max and where is min

{{{f}}}''{{{(x) =12x+6}}}


{{{f}}}''{{{(-5) =12(-5)+6=-54}}} => negative, the absolute maximum is at{{{ -5}}}


{{{f}}}''{{{(4) =12(4)+6=54}}} => positive, the absolute minimum is at {{{4}}}



{{{ graph( 600, 600, -15, 15, -500, 500, 2x^3+3x^2-120x+2) }}}