Is #f(x) = 8x^4+3x^3 + 4x^2+x-4# concave or convex at #x=11#?

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Answer 1 · 2018-02-01T16:12:57

Concave

The concavity of a function is given by it's third derivative:
When #f''(x)>0#, the function is concave and when #f''(x)<0#, it's convex.

Let's determine the first derivative:

#f'(x)=(8x^4+3x^3+4x^2+x-4)'#

Given #(u+v)'=u'+v'#, #(ku)'=ktimesu'# and #(x^n)'=ntimesx^(n-1#:

#f'(x)=32x^3+9x^2+2x+1#

Now, determine the second derivative:

#f''(x)=(32x^3+9x^2+2x+1)'#

#f''(x)=96x^2+18x+2#

Plug in #x=11#:

#f''(11)=96times11^2+18times11+2=11.816#

As #f''(11)>0#, the function is #concave# at #x=11#