Question 1155428

Rolle's Theorem
Let {{{f}}} be differentiable on the open interval ({{{a}}},{{{b}}}) and continuous on the closed interval [{{{a}}},{{{b}}}]. 
Then if {{{f(a)=f(b)}}}, then there is {{{at}}}{{{ least}}} one point {{{c}}} in ({{{a}}},{{{b}}}) where {{{f}}}'{{{(c)=0}}}.



{{{f(x)=x(x-8)^2}}}; 

[{{{0}}},{{{8}}}]=[{{{a}}},{{{b}}}]

check if {{{f(a)=f(b)}}} for

{{{a=0}}}
{{{b=8}}}

{{{f(0)=0(0-8)^2=0}}}
{{{f(8)=8(8-8)^2=0}}}

=>  {{{f(a)=f(b)}}}

then derivate

{{{f(x)=x(x-8)^2}}}

apply the Product Rule : {{{(f*g )}}}'={{{f}}}'*{{{g}}}+{{{f}}}* {{{g}}}'

{{{f}}}'{{{(x)=(d/dx)x*(x-8)^2+x*(d/dx)(x-8)^2}}}

{{{f}}}'{{{(x)=1*(x-8)^2+x*2(x-8)}}}

{{{f}}}'{{{(x)=x^2-16x+64+2x^2-16x}}}

{{{f}}}'{{{(x) = 3 x^2 - 32 x + 64}}}

find  point {{{c}}} in ({{{a}}},{{{b}}}) where {{{f}}}'{{{(c)=0}}}

{{{ 3x^2 - 32x + 64=0}}}

 {{{3x^2 - 24x-8x + 64=0}}}

 {{{(3 x^2 - 24x)-(8x - 64)=0}}}

{{{(x - 8) (3x - 8) = 0}}}

solutions:

if {{{(x - 8) = 0}}}=>{{{x=8}}}
if {{{ (3x - 8) = 0}}}=>{{{x=8/3}}}

 the point(s) that are guaranteed to exist by Rolle's theorem are:

({{{8}}},{{{0}}}) and ({{{8/3}}},{{{0}}})