# What are the points of inflection, if any, of f(x)=2x^3+x^2+x+3 ?

##### 1 Answer
Mar 9, 2016

The point of inflection occurs at $x = - \frac{1}{6}$

#### Explanation:

The first derivative gives us the gradient of the line for a particular value of $x$:

graph{(2x^3+x^2+x+3-y)(6x^2+2x+1-y)=0 [-10, 10, -5, 5]}
$f \left(x\right) = 2 {x}^{3} + {x}^{2} + x + 3$
$f ' \left(x\right) = 6 {x}^{2} + 2 x + 1$

The second derivative tells us whether the gradient is increasing or decreasing.

This is shown as the parabola on the graph.

When the second derivative is positive, the function is concave upward. This looks like an open parabola facing upwards.

When the second derivative is negative, the function is concave downward. This looks like an open parabola facing downwards.

The inflection point is where the function goes from concave upwards to concave downwards (or vice versa).

$f ' ' \left(x\right) = 12 x + 2$

So you can an see that $\left(12 x + 2\right)$ is -ve up to $x = - \frac{1}{6}$ and +ve thereafter.

So $f \left(x\right)$ is "concave down" up to $x = - \frac{1}{6}$

and $f \left(x\right)$ is "concave up" from $x = - \frac{1}{6}$

So the point of inflection occurs at $x = - \frac{1}{6}$