Question 1155976

{{{f(x)=-e^x(x-3)}}}

Derivative:

{{{d/dx(-e^x (x - 3)) = -e^x (x - 2)}}}

=> critical points: 

{{{-e^x (x - 2)=0}}} => only if  {{{(x - 2)=0}}} =>{{{x=2}}}

second derivative:

{{{f}}}'{{{(x) = -e^x (x - 1)}}}..........now, plug the three critical number {{{x=2}}} into the second derivative 

{{{f}}}'{{{(x) = -e^2 (2 - 1)}}}

{{{f}}}'{{{(x) = -e^2 (1)}}}

{{{f}}}'{{{(x) = -e^2 }}}


At  {{{2}}}, the second derivative is {{{negative}}} ( {{{-e^2 }}}). This tells you that {{{f}}} is concave down where {{{x=2}}}, and therefore that there’s a local{{{ max}}} at {{{ 2}}}.


{{{ graph( 600, 600, -10, 10, -10, 10, -e^x*(x-3)) }}}