# The function 3x^(3)+6x^(2)+6x+10 is maxima, minima or point of inflection?

Feb 12, 2018
• No mins or maxes
• Point of Inflection at $x = - \frac{2}{3}$.

graph{3x^3 + 6x^2 + 6x + 10 [-10, 10, -10, 20]}

## Mins and Maxes

For a given $x$-value (let's call it $c$) to be a max or min for a given function, it has to satisfy the following:

$f ' \left(c\right) = 0$ or undefined.

These values of $c$ are also called your critical points.

Note: Not all critical points are max/mins, but all max/mins are are critical points

So, let's find these for your function:

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

$\implies \frac{d}{\mathrm{dx}} \left(3 {x}^{3} + 6 {x}^{2} + 6 x + 10\right) = 0$

$\implies 9 {x}^{2} + 12 x + 6 = 0$

This doesn't factor, so let's try quadratic formula:

$x = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \left(9\right) \left(6\right)}}{2 \left(9\right)}$

$\implies \frac{- 12 \pm \sqrt{- 72}}{18}$

...and we can stop right there. As you can see, we end up having a negative number under the square root. Hence, there are no real critical points for this function.

-

## Points of Inflection

Now, let's find points of inflection. These are points where the graphÂ has a change in concavity (or curvature). For a point (call it $c$) to be a point of inflection, it must satisfy the following:

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

Note: Not all such points are points of inflection, but all points of inflection must satisfy this.

So let's find these:

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

$\implies \frac{d}{\mathrm{dx}} \left(\frac{d}{\mathrm{dx}} \left(3 {x}^{3} + 6 {x}^{2} + 6 x + 10\right)\right) = 0$

$\implies \frac{d}{\mathrm{dx}} \left(9 {x}^{2} + 12 x + 6 = 0\right)$

$\implies 18 x + 12 = 0$

$\implies x = - \frac{12}{18} = - \frac{2}{3}$

Now, we need to check if this is in fact a point of inflection. So we'll need to verify that $f ' ' \left(x\right)$ does in fact switch sign at $x = - \frac{2}{3}$.

So let's test values to the right & left of $x = - \frac{2}{3}$:

Right:
$x = 0$
$f ' ' \left(0\right) = 12$

Left:
$x = - 1$
$f ' ' \left(- 1\right) = - 6$

We don't care as much what the actual values are, but as we can clearly see, there's a positive number to the right of $x = - \frac{2}{3}$, and a negative number to the left of $x = - \frac{2}{3}$. Hence, it is indeed a point of inflection.

To summarize, $f \left(x\right)$ has no critical points (or mins or maxes), but it does have a point of inflection at $x = - \frac{2}{3}$.

Let's take a look at the graph of $f \left(x\right)$ and see what these results mean:

graph{3x^3 + 6x^2 + 6x + 10 [-10, 10, -10, 20]}

This graph is increasing everywhere, so it doesn't have any place where the derivative = 0. However, it does go from curved down (concave down) to curved up (concave up) at $x = - \frac{2}{3}$.

Hope that helped :)