Is #f(x) =(2x+8)^3/(x^2/3+1)# concave or convex at #x=-4#?

1 Answer
Apr 4, 2018

Below

Explanation:

#f(x)=(2x+8)^3/(x^2/3+1)#
#f'(x)=((x^2/3+1)times6(2x+8)^2-(2x+8)^3((2x)/3))/(x^2/3+1)^2#

For stationary points, #f'(x)=0#

ie #(x^2/3+1)times6(2x+8)^2-(2x+8)^3((2x)/3)=0#

#(2x+8)^2(6(x^2/3+1)-(2x+8)(2x/3))=0#

#(2x+8)^2(2x^2+6-(4x^2)/3-16x/3)=0#

#(2x+8)^2((2x^2)/3+6-(16x)/3)=0#

#(2x+8)^2=0# OR #(2x^2)/3+6-(16x)/3=0#

Since you are finding #x=-4#

You can use a table

when
#x=-3#, #f'(x)=7#
#x=-4#, #f'(x)=0#
#x=-5#, #f'(x)=111/49#

Therefore, the point at x=-4 is neither concave up nor concave down. Instead at x=4, you will have something that looks like this cubic graph at x=0

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  • (i don't remember its name so I had to give you an example instead)

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graph{x^3 [-10, 10, -5, 5]}