# What are the points of inflection, if any, of f(x)=x^4-x^3+6 ?

Feb 17, 2018

See explanation.

#### Explanation:

Point of inflection can be calculated as the zero(s) of the second derivative. Here we have:

$f \left(x\right) = {x}^{4} - {x}^{3} + 6$

$f ' \left(x\right) = 4 {x}^{3} - 3 {x}^{2}$

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

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

If we look at the graph of $f ' ' \left(x\right)$ we see:

graph{12x^2-6x [-3.077, 3.08, -1.538, 1.54]}

The derrivative changes signs at both $x = 0$ and $x = \frac{1}{2}$, so:

Answer: The function has 2 points of inflection: $x = 0$ and $x = \frac{1}{2}$

Feb 17, 2018

$x = 0 , x = \frac{1}{2}$

#### Explanation:

The points of inflection are when the graph of $f \left(x\right)$ changes concavity. For example, if $f \left(x\right)$ is concave up from $\left(- \infty , 3\right)$ and concave down from $\left(3 , \infty\right)$, then we know that the point of inflection is at $3$ because that is where $f \left(x\right)$ changes concavity.

The way to find whether $f \left(x\right)$ is concave up or concave down is to take the second derivative of it. The first derivative determines if it is increasing or decreasing, and the second derivative determines if it is concave up or concave down. If the second derivative at a point is positive, then $f \left(x\right)$ is concave up, but if it is negative, then $f \left(x\right)$ is concave down.

To take the second derivative of a function, just derive it twice.

$f \left(x\right) = {x}^{4} - {x}^{3} + 6$

$f ' \left(x\right) = 4 {x}^{3} - 3 {x}^{2}$ (using the power rule)

$f ' ' \left(x\right) = 12 {x}^{2} - 6 x$ (using the power rule again)

By the way, power rule states that the derivative of a function such as $C {x}^{n}$ (where $n$ is an integer and $C$ is a constant) is equal to $n C {x}^{n - 1}$.

Now that we have the second derivative $f ' ' \left(x\right) = 12 {x}^{2} - 6 x$, we need to find out when it changes signs. We know that whenever a function changes signs, it crosses $0$, so we must find out when $f ' ' \left(x\right) = 0$.

$12 {x}^{2} - 6 x = 0$

$6 x \left(2 x - 1\right) = 0$ (by factoring)

$x = 0 , x = \frac{1}{2}$

Therefore, we know know that the points of inflection are at $x = 0$ and at $x = \frac{1}{2}$.

This is the graph of $f \left(x\right)$.
graph{x^4 - x^3 + 6 [-7.194, 8.616, -0.42, 7.48]}

This is the graph of $f ' \left(x\right)$.
graph{4x^3 - 3x^2 [-9.29, 10.72, -4.11, 5.885]}

This is the graph of $f ' ' \left(x\right)$.
graph{12x^2 - 6x [-9.29, 10.72, -4.11, 5.885]}