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2023-02-27

What are the points of inflection, if any, of $f\left(x\right)={x}^{4}-{x}^{3}+6$? Gerald Dickerson

Point of inflection can be calculated as the zero(s) of the second derivative. Here we have:
$f\left(x\right)={x}^{4}-{x}^{3}+6$
$f\prime \left(x\right)=4{x}^{3}-3{x}^{2}$
$f\prime \prime \left(x\right)=12{x}^{2}-6x$
$f\prime \prime \left(x\right)=0⇔6x\left(2x-1\right)=0⇔x=0\vee x=\frac{1}{2}$
The derrivative changes signs at both $x=0$ and $x=\frac{1}{2}$, therefore:
Answer: The function has 2 points of inflection: $x=0$ and $x=\frac{1}{2}$ Aubree Phelps

The points of inflection are when the graph of f(x) changes concavity. For example, if f(x) 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(x) changes concavity.
Take the second derivative of f(x) to determine whether it is concave up or concave down. The first derivative indicates whether it is increasing or decreasing, while the second derivative indicates whether it is concave up or concave down. If the second derivative at a point is positive, f(x) is concave up; otherwise, f(x) is concave down.
Simply derive the second derivative of a function twice.
$f\left(x\right)={x}^{4}-{x}^{3}+6$
$f\prime \left(x\right)=4{x}^{3}-3{x}^{2}$ (using the power rule)
$f\prime \prime \left(x\right)=12{x}^{2}-6x$ (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 $nC{x}^{n-1}$.
Now that we have the second derivative $f\prime \prime \left(x\right)=12{x}^{2}-6x$, 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 $f\prime \prime \left(x\right)=0$.
$12{x}^{2}-6x=0$
$6x\left(2x-1\right)=0$ (by factoring)
$x=0,x=\frac{1}{2}$
Hence, we know know that the points of inflection are at $x=0$ and at $x=\frac{1}{2}$.

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