# How do you find Find the points of inflection and discuss the concavity of the function f(x) = 2x^3-3x^2-12x+5?

Apr 3, 2015

Concavity and inflection points are investigated using the second derivative and its sign.

$f \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 12 x + 5$

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

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

Now, $f ' ' \left(x\right)$ is continuous, so if it changes sign it must be $0$ at some point (Intermediate Value Theorem).
(A discontinuous function can also change sign at a discontinuity.)

$f ' ' \left(x\right) = 12 x - 6 = 0$ at $x = \frac{1}{2}$

Left of $\frac{1}{2}$ (say at $x = 0$, we see that $f ' ' \left(x\right) < 0$ So the graph of $f$ is concave down to the left of $x = \frac{1}{2}$

If $x$ is right of $\frac{1}{2}$, (say at $x = 10$) then $f ' ' \left(x\right) > 0$ so the graph of $f$ is concave up to the right of $x = \frac{1}{2}$.

The concavity changes as we cross $x = \frac{1}{2}$ if there is a point on the graph at $x = \frac{1}{2}$, (if $\frac{1}{2}$ is in the domain of $f$) then that point is an inflection point.

$f \left(\frac{1}{2}\right) = 2 {\left(\frac{1}{2}\right)}^{3} - 3 {\left(\frac{1}{2}\right)}^{2} - 12 \left(\frac{1}{2}\right) + 5 = \frac{1}{4} - \frac{3}{4} - 6 + 5 = - \frac{2}{4} - 1 = - \frac{3}{2}$

So the point $\left(\frac{1}{2} , \frac{3}{2}\right)$ is an inflection point.