# Is f(x)=-2x^5-2x^3+3x^2-x+3 concave or convex at x=-1?

Jan 29, 2016

Convex.

#### Explanation:

You can tell if a function is concave or convex by the sign of its second derivative:

• If $f ' ' \left(- 1\right) < 0$, then $f \left(x\right)$ is concave at $x = - 1$.
• If $f ' ' \left(- 1\right) > 0$, then $f \left(x\right)$ is convex at $x = - 1$.

To find the second derivative, apply the power rule to each term twice.

$f \left(x\right) = - 2 {x}^{5} - 2 {x}^{3} + 3 {x}^{2} - x + 3$

$f ' \left(x\right) = - 10 {x}^{4} - 6 {x}^{2} + 6 x - 1$

$f ' ' \left(x\right) = - 40 {x}^{3} - 12 x + 6$

Find the sign of the second derivative at $x = - 1 :$

$f ' ' \left(- 1\right) = - 40 {\left(- 1\right)}^{3} - 12 \left(- 1\right) + 6$

This mostly becomes a test of keeping track of your positives and negatives.

$f ' ' \left(- 1\right) = - 40 \left(- 1\right) + 12 + 6 = 40 + 18 = 58$

Since this is $> 0$, the function is convex at $x = - 1$. Convexity on a graph is characterized by a $\cup$ shape.

We can check the graph of the original function:

graph{-2x^5-2x^3+3x^2-x+3 [-2.5, 2, -30, 30]}