Is #f(x)=3x^5-x^3+8x^2-x# concave or convex at #x=0#?

Question

1260 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-11T03:07:28

#f(x)# is concave at #x=0.#

#f(x)# is concave at #x=a# if #f''(a)>0#.

#f(x)# is convex at #x=a# if #f''(a)>0#.

Here, #f(x)=3x^5-x^3+8x^2-x, a=0.# Knowing that, let's take the second derivative.

#f'(x)=15x^4-3x^2+16x-1#

#f''(x)=20x^3-6x+16#

Evaluate at #0:#

#f''(0)=20(0^3)-6(0)+16=16>0#

#f(x)# is concave at #x=0.#