# For what values of x is f(x)= 7x^3 + 2 x^2 + 7x -2  concave or convex?

Mar 26, 2017

Concave down for $x < - \frac{2}{21}$, concave up for $x > - \frac{2}{21}$

#### Explanation:

First, we can try to find inflection points for this function. An inflection point is a point where the concavity changes, so finding this point is often helpful when analyzing concavity.

The process is straightforward; we will find $x$ such that ${d}^{2} / {\mathrm{dx}}^{2} f \left(x\right) = 0$, that is, when the second derivative of $f$ is zero.

Start by finding the 1st derivative, by simply applying the power rule to each term:

$\frac{d}{\mathrm{dx}} f \left(x\right) = 21 {x}^{2} + 4 x + 7$

Then, differentiate again to find the 2nd derivative:

${d}^{2} / {\mathrm{dx}}^{2} f \left(x\right) = 42 x + 4$

So, now we set the thing equal to zero:

$0 = 42 x + 4$

And solve for $x$:

$x = - \frac{2}{21}$

So now we know that the function has one inflection point. What about the rest of possible values for $x$?

Well, if a segment of a graph is concave up (its slope is increasing) then the 2nd derivative will be positive. And if a segment is concave down, with a decreasing slope, the 2nd derivative will be negative.

Since we know that the 2nd derivative switches from negative to positive or vice versa at $x = - \frac{2}{21}$, let's see whether it's positive or negative, at, for instance, $x = 0$:

${d}^{2} / {\mathrm{dx}}^{2} f \left(0\right) = 42 \cdot 0 + 4 = 4$

Interesting. So the 2nd derivative is positive at $x = 0$. This tells us that, since the only switch occurs at $x = - \frac{2}{21}$, all $x > - \frac{2}{21}$ have a positive 2nd derivative as well, and are therefore concave up.

On the other hand, all $x < - \frac{2}{21}$ are concave down, since the switch must occur at $x = - \frac{2}{21}$.

Hopefully this makes sense.