# Is f(x)=x^4-4x^3+x-4 concave or convex at x=-1?

Mar 22, 2016

convex

#### Explanation:

To find that out, we need to get the second derivative first.

Getting the first derivative.

$\left[1\right] \text{ } f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{4} - 4 {x}^{3} + x - 4\right)$

We can easily find this using power rule.

$\left[2\right] \text{ } f ' \left(x\right) = 4 {x}^{3} - 12 {x}^{2} + 1$

Getting the second derivative.

$\left[1\right] \text{ } f ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(4 {x}^{3} - 12 {x}^{2} + 1\right)$

Use power rule again.

$\left[2\right] \text{ } f ' ' \left(x\right) = 12 {x}^{2} - 24 x$

Now that we know the second derivative, we will evaluate $f ' ' \left(x\right)$ at $x = - 1$ to check its concavity.

• If $f ' ' \left(x\right) > 0$, then it is concave up or convex
• If $f ' ' \left(x\right) < 0$, then it is concave down or concave

$\left[1\right] \text{ } f ' ' \left(- 1\right) = 12 {\left(- 1\right)}^{2} - 24 \left(- 1\right)$

$\left[2\right] \text{ } f ' ' \left(- 1\right) = 12 + 24$

$\left[3\right] \text{ } f ' ' \left(- 1\right) = 36$

Since $f \left(x\right)$ is 36 at $x = - 1$ and 36 is greater than 0, then $f \left(x\right)$ is convex at $x = - 1$.