# How would you determine where the function f(x) is concave up and concave down for f(x)=2x^3-3x^2-12x+1?

May 22, 2015

First, I would find the vertexes. Then, the inflection point.

The vertexes indicate where the slope of your function change, while the inflection points determine when a function changes from concave to convex (and vice-versa).

In order to find the vertexes (also named "points of maximum and minimum"), we must equal the first derivative of the function to zero, while to find the inflection points we must equal the second derivative to zero.

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} - 6 x - 12 = 0$

$\frac{6 \pm \sqrt{36 - 4 \left(6\right) \left(- 12\right)}}{12}$
$\frac{6 \pm 18}{12}$
${x}_{1} = 2$
${x}_{2} = - 1$

For $f \left(2\right) = 2 {\left(2\right)}^{3} - 3 {\left(2\right)}^{2} - 12 \left(2\right) + 1 = - 19$
and $f \left(- 1\right) = 2 {\left(- 1\right)}^{3} - 3 {\left(- 1\right)}^{2} - 12 \left(- 1\right) + 1 = 8$

So, your vertexes are ${V}_{1} \left(2 , - 19\right)$ and ${V}_{2} \left(- 1 , 8\right)$

Now, let's see the behaviour of your function relating to concavity.

$f ' ' \left(x\right) = \frac{{\mathrm{dy}}^{2}}{{d}^{2} x} = 12 x - 6 = 0$

$12 x - 6 = 0$
$12 x = 6$
$x = \frac{1}{2}$

For $f \left(\frac{1}{2}\right) = 2 {\left(\frac{1}{2}\right)}^{3} - 3 {\left(\frac{1}{2}\right)}^{2} - 12 \left(\frac{1}{2}\right) + 1$
$f \left(\frac{1}{2}\right) = \frac{1}{4} - \frac{3}{4} - 6 + 1 = - \frac{11}{2}$

Thus, your inflection point is $\left(\frac{1}{2} , - \frac{11}{2}\right)$

About concavity up or down: we just have one more thing to know about these vertexes, which is the way the function is taking after the vertex - "reading" it from left to right.

This is done by substituting the values we found for $x$ in $f ' \left(x\right)$ in the function $f ' ' \left(x\right)$:

$f ' ' \left(2\right) = 12 \left(2\right) - 6 = 18$ (which indicates point of minimum, as $f ' ' \left(x\right) > 0$)
$f ' ' \left(- 1\right) = 12 \left(- 1\right) - 6 = - 18$ (which indicates point of maximum, as $f ' ' \left(x\right) < 0$)

Check the graph:

graph{2x^3-3x^2-12x+1 [-30.15, 32.28, -19.86, 11.34]}

May 23, 2015

$f \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 12 x + 1$

To investigate concavity, we need to look at the sign of $f ' ' \left(x\right)$.

$f ' \left(x\right) = 6 {x}^{2} - 6 x - 12$

$f ' ' \left(x\right) = 12 x - 6$

In general a function can change signs by either crossing the $x$ axis (having a value of $0$), or by "teleporting" across the $x$ axis (having a discontinuity).

In the case, there are no discontinuities, so the only place $f ' '$ might change signs is at:

$12 x - 6 = 0$ so $x = \frac{1}{2}$.
Now we check the sign of $f ' '$ (see note below)

$\left.\begin{matrix}\text{interval " & " test number " & "sign of "f'' \\ (-oo1/2) & color(white)"sssssss"0 & 12(0)-6 " is "- \\ (1/21oo) & color(white)"sssssss"1 & 12(1)-6 " is } +\end{matrix}\right.$

$f$ is concave down on $\left(- \infty , \frac{1}{2}\right)$ and concave up on $\left(\frac{1}{2.} \infty\right)$

(This was not asked, but $f$ has inflection point $\left(\frac{1}{2} , f \left(\frac{1}{2}\right)\right)$)

Note
I prefer the general approach to finding the sign of $f ' '$ . This involves cutting the number line into intervals and testing the sign of $f ' '$ on each interval.
There are other methods for linear and quadratic $f ' '$, many won't work for more general functions.

For example, in this problem $f ' ' \left(x\right) = 12 x - 6$ is a line with positive slope, so $y$ is negative on the left of (and positive on the right right of) the $x$ intercept, $\frac{1}{2}$.