Question 1155999
{{{f(x) = matrix(2,1,"",x*(e^(-x^2/98)))}}}

[{{{-4}}},{{{14}}}]


derivate to find critical points:


{{{f}}}'{{{(x) = -(1/49)*matrix(2,1,"",e^(-x^2/98))*(x^2 - 49)}}}


equal to zero:



 {{{-(1/49)* matrix(2,1,"",e^(-x^2/98)) *(x^2 - 49)=0}}} ......since  will be equal to zero if


{{{matrix(2,1,"",e^(-x^2/98)) =0}}} or {{{(x^2 - 49)=0}}}


{{{matrix(2,1,"",e^(-x^2/98) ) =0}}}=> solution does not exist


{{{(x^2 - 49)=0}}}

{{{(x - 7)(x+7)=0}}}

=>{{{x=7}}} or {{{x=-7}}}


{{{f(7) = matrix(2,1,"",7(e^(-7^2/98)))}}}

{{{f(7) = matrix(2,1,"",7(e^(-49/98)))}}}

{{{f(7) = matrix(2,1,"",7(e^(-1/2)))}}}

{{{f(7) = 7sqrt(e)}}}


{{{f(7) = 11.54}}}



{{{f(-7) = matrix(2,1,"",-7(e^(-(-7)^2/98)))}}}

{{{f(-7) = matrix(2,1,"",-7(e^(-49/98)))}}}

{{{f(-7) = matrix(2,1,"",-7(e^(-1/2)))}}}

{{{f(-7) = -7sqrt(e)}}}

{{{f(7) = -11.54}}}


both solutions are in given interval

use second derivative test to determine at what point is maximum and minimum:

{{{f}}}''{{{(x) =(d/dx)(-(1/49)* matrix(2,1,"",(e^(-x^2/98))* (x^2 - 49))) = matrix(2,1,"",(e^(-x^2/98))* x* (x^2 - 147))/2401}}}


if {{{x=7}}}


{{{matrix(2,1,"",(e^(-7^2/98)) *7 (-147 + 7^2))/2401=-0.1732944742}}}=> negative, so at {{{x=7}}} is a {{{max}}}


if {{{x=-7}}}

{{{matrix(2,1,"",(e^(-(-7)^2/98)) *-7 (-147 + (-7)^2))/2401=0.1732944742}}}=> positive, so at {{{x=-7}}} is a  min



{{{ graph( 600, 600, -10, 10, -10, 10, x*(e^(-x^2/98))) }}}