Question 1170622
{{{f(x) = 3x^5 - 5x^3}}} defined on [{{{-3}}}, {{{3}}}]

Identify the optimal point(s) of this function.

Find the global max and min points of this function.

{{{x}}}-intercepts:

{{{0 = (3x^2 - 5)x^3}}}-> {{{x=0}}} or
{{{0 = (3x^2 - 5)}}} ->{{{5 = (3x^2 - 5)}}} ->{{{5/3 = x^2}}}  -> {{{x=+-sqrt(5/3) }}}

{{{x}}}-intercepts are at: ({{{sqrt(5/3 )}}},{{{0}}}) and ({{{-sqrt(5/3 )}}},{{{0}}})

{{{y}}}-intercept:

{{{f(x) = 3*0^5 - 5*0^3=0}}}

{{{y}}}-intercept at: ({{{0}}},{{{0}}})

global max : 

use first derivative test
 
{{{(d/dx)(3x^5 - 5x^3) =5*3x^4-3*5x^2}}}

{{{(d/dx)(3x^5 - 5x^3) =15x^4-15x^2}}}

{{{(d/dx)(3x^5 - 5x^3) = 15x^2(x^2 - 1)}}}

->{{{(x^2 - 1)=0}}} ->{{{x^2 = 1}}}->{{{x=1}}} or {{{x=-1}}}


then

{{{f(x) = 3*1^5 - 5*1^3=-2}}}

{{{f(x) = 3*(-1)^5 - 5*(-1)^3=-3-(-5)=-3+5=2}}}



max{ {{{ 3x^5 - 5x^3}}} } ={{{ 2}}} at{{{x = -1}}}
and 
min { {{{3x^5 - 5x^3}}} } ={{{ -2}}} at {{{x = 1}}}

use given interval as critical points:

{{{f(-3) = 3*(-3)^5 - 5*(-3)^3=-594}}}
min ({{{-3}}},{{{ -594}}} )

{{{f(3) = 3*(3)^5 - 5*(3)^3=594}}}
max({{{3}}},{{{594}}})