# How do you find the maximum, minimum and inflection points and concavity for the function f(x)=6x^6+12x+6?

Jun 20, 2016

Relative minimum at $x = {\left(- \frac{1}{3}\right)}^{\frac{1}{5}}$.
Always concave up and no inflection points.

#### Explanation:

$f ' \left(x\right) = 36 {x}^{5} + 12$
$f ' ' \left(x\right) = 180$

To find relative maxima, set $f ' \left(x\right) = 0$.
So, $36 {x}^{5} + 12 = 0$
$\therefore$ ${x}^{5} = - \frac{1}{3}$
$\therefore x = {\left(- \frac{1}{3}\right)}^{\frac{1}{5}}$

Now, use the sign test to determine the relative maximum and minimum of the function.

For - infinity$< x < {\left(- \frac{1}{3}\right)}^{\frac{1}{5}}$, the function is decreasing.
For ${\left(- \frac{1}{3}\right)}^{\frac{1}{5}} < x <$ infinity, the function is increasing.
Therefore, the function has relative minimum at $x = {\left(- \frac{1}{3}\right)}^{\frac{1}{5}}$

Regarding concavity of the function, since the second derivative is always positive, we can determine that the function is always concave up and has no inflection points.