Question 1137124


Sketch the graph of the function

{{{f(x)=2x^3-x^2-8x+4}}}

Solution:

1. Find the relative max and min 

{{{f}}}′={{{6x^2-2x-8=2(3x^2-x-4)=2(x + 1) (3x - 4)}}}

critical numbers:
{{{x=-1}}}
{{{x=4/3}}}


{{{f}}}′′={{{12x-2}}}


evaluate at the critical numbers

{{{f}}}′′{{{(-1)=-12-2=-14<0}}}=>  f concave down

relative max at {{{x= -1}}}

{{{f(-1)=2(-1)^3-(-1)^2-8(-1)+4=9}}}


=> point ({{{-1}}},{{{ 9}}}) is a relative max


{{{f}}}′′{{{(4/3)=12(4/3)-2=14 >0}}} f concave up

relative min at {{{x=4/3}}}

{{{f(4/3)=2(4/3)^3-(4/3)^2-8(4/3)+4=-3.7}}}

 =>point:({{{4/3}}}, {{{-3.7}}}) is a relative min


{{{drawing( 600,600, -10,10, -10, 10,

circle(4/3,-3.7,.12),locate(4/3,-3.7,min(4/3,-3.7)),
circle(-1,9,.12),locate(-1,9,max(-1,9)),

 graph( 600,600, -10,10, -10, 10, 2x^3-x^2-8x+4)) }}}