Answer:
Relative maximum at x=12
Relative minimum at x=-4
Inflection point at x=4
Step-by-step explanation:
To find all the critical points of the function we need the first derivative of the function
[tex]y(x)=-2x^3+24x^2+288x-35\\y'(x)=-2(3)x^{3-1}+24(2)x^{2-1}+288(1)\\y'(x)=-6x^2+48x+288[/tex]
According to Fermat's theorem, it will be a critical point in y'(x)=0, then
[tex]y'(x)=-6x^2+48x+288\\0=-6x^2+48x+288\\\\-6x^2+48x+288=0\\-6(x^2-8x-48=0\\x^2-8x-48=0\\(x-12)(x+4)=0\\x_1=12\\x_2=-4\\[/tex]
With the test of the second derivative, we can know if the points are relative extrema or inflections.
[tex]y'(x)=-6x^2+48x+288\\y''(x)=-6(2)x^{2-1}+48(1)\\y''(x)=-12x+48[/tex][tex]y''(12)=-12(12)+48\\y''(12)=-144+48=-96[/tex]
[tex]y''(-4)=-12(-4)+48\\y''(-4)=48+48=96[/tex]
The second derivative test says: if [tex]f''(x)<0[/tex] then x is a relative maximum, if [tex]f''(x)>0[/tex] then x is a relative minimum. So x=12 is a relative maximum and x=-4 is a relative minimum.
To find the inflection point y''(x)=0
[tex]y''(x)=-12x+48\\0=-12x+48\\12x=48\\x=48/12=4[/tex]
