# How do you determine whether the function f(x)=(2x-1)^2(x-3)^2 is concave up or concave down?

Sep 11, 2015

To determine where the graph of $f$ is concave up and where it is concave down, look at the sign of $f ' ' \left(x\right)$.

#### Explanation:

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

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

$= 2 \left(2 x - 1\right) \left(x - 3\right) \left[2 \left(x - 3\right) + \left(2 x - 1\right)\right]$

$= 2 \left(2 x - 1\right) \left(x - 3\right) \left(4 x - 7\right)$

$f ' ' \left(x\right) = 2 \left(2\right) \left(x - 3\right) \left(4 x - 7\right)$

$+ 2 \left(2 x - 1\right) \left(1\right) \left(4 x - 7\right)$

$+ 2 \left(2 x - 1\right) \left(x - 3\right) \left(4\right)$

$= 48 {x}^{2} - 168 x + 61$

Set $f ' ' \left(x\right) = 0$ to find partition numbers: $\frac{21 \pm 5 \sqrt{3}}{12}$

Test each interval:

On $\left(- \infty , \frac{21 - 5 \sqrt{3}}{12}\right)$, $f ' '$ is positive, graph is concave up

on $\frac{21 - 5 \sqrt{3}}{12} , \frac{21 + 5 \sqrt{3}}{12}$, $f ' '$ is negative, graph is concave down

on $\left(\frac{21 + 5 \sqrt{3}}{12} , \infty\right)$, $f ' '$ is positive, graph is concave up