# For what values of x is f(x)=-x^3+x^2+2x+5 concave or convex?

Jul 23, 2017

$f \left(x\right)$ is concave up (convex) from $\left(- \infty , \frac{1}{3}\right)$
$f \left(x\right)$ is concave down from $\left(\frac{1}{3} , \infty\right)$

#### Explanation:

To find where the equation is concave and convex we need to first find the first derivative using the power rule.

$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

The first derivative is:

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

Now we find the second derivative:

$f ' ' \left(x\right) = - 6 x + 2$

We set it equal to zero and solve:

$- 6 x + 2 = 0$

$x = \frac{1}{3}$

Now we create a test interval from $\left(- \infty , \frac{1}{3}\right)$ and $\left(\frac{1}{3} , \infty\right)$.
Then we pick a number from the left and right and plug it into our

second derivative $f ' ' \left(x\right) = - 6 x + 2$.

Keep in mind that it can be any number I picked $0$ which is a number from the left of $\frac{1}{3}$, then I selected $1$ which is a number from the right of $\frac{1}{3}$.

Now if the number you get after plugging it in is positive then it means for that interval the graph is concave up. If the number is negative then the graph is concave down for that interval only.

So we can say that for $f ' ' \left(0\right) = 2$ the graph is concave up.
Now for $f ' ' \left(1\right) = - 4$ the graph is concave down.

$f \left(x\right)$ is concave up from $\left(- \infty , \frac{1}{3}\right)$
$f \left(x\right)$ is concave down from $\left(\frac{1}{3} , \infty\right)$