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

Jan 17, 2016

Neither. It is a point of inflection.

#### Explanation:

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

• If $f ' ' \left(0\right) > 0$, then $f \left(x\right)$ is convex when $x = 0$.
• If $f ' ' \left(0\right) < 0$, then $f \left(x\right)$ is concave when $x = 0$.

Find the function's second derivative.

$f \left(x\right) = - 5 {x}^{5} - 2 {x}^{4} - 2 {x}^{3} + 14 x - 17$
$f ' \left(x\right) = - 25 {x}^{4} - 8 {x}^{3} - 6 {x}^{2} + 14$
$f ' ' \left(x\right) = - 100 {x}^{3} - 24 {x}^{2} - 12 x$

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

$f ' ' \left(0\right) = 0$

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 $x = 0$.