Is #f(x)=-5x^5-2x^4-2x^3+14x-17# concave or convex at #x=0#?

1 Answer
mason m
Jan 17, 2016

Neither. It is a point of inflection.

Explanation:

Convexity and concavity are determined by the sign of the second derivative.

  • If #f''(0)>0#, then #f(x)# is convex when #x=0#.
  • If #f''(0)<0#, then #f(x)# is concave when #x=0#.

Find the function's second derivative.

#f(x)=-5x^5-2x^4-2x^3+14x-17#
#f'(x)=-25x^4-8x^3-6x^2+14#
#f''(x)=-100x^3-24x^2-12x#

Find the sign of the second derivative at #x=0#.

#f''(0)=0#

Notice that the sign of the second derivative is neither positive nor negative. This means that the function is neither convex nor concave. This means that is may be a point of inflection.

We can check a graph of the function:

graph{-5x^5-2x^4-2x^3+14x-17 [-2.5, 2.5, -120, 100]}

Graphically, #x=0# does appear to be a point of inflection (the concavity shifts). This is testable by seeing if the sign of the second derivative goes from positive to negative or vice versa around the point #x=0#.

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