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

Question

Emilia Aguilar · Accepted Answer

Neither. It is a point of inflection.

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

Luke Alvarez · Answer

undefined To determine the concavity of $ f(x) = -5x^5 - 2x^4 - 2x^3 + 14x - 17 $ at $ x = 0 $, we need to examine the second derivative $ f''(x) $ at that point.

First, find the first derivative $ f'(x) $:
$$ f'(x) = -25x^4 - 8x^3 - 6x^2 + 14 $$

Next, find the second derivative $ f''(x) $:
$$ f''(x) = -100x^3 - 24x^2 - 12x $$

Now, evaluate $ f''(0) $:
$$ f''(0) = -12(0) = 0 $$

Since $ f''(0) = 0 $, we cannot determine the concavity at $ x = 0 $ solely from the second derivative test. We need to further analyze the function using other methods, such as the first derivative test or plotting the function on a graph.