Question

How do you find the inflection points for $f\left( x \right)={{x}^{4}}-10{{x}^{3}}+24{{x}^{2}}+3x+5?$

Answer 1

Hint: The inflection points of a curve is the point at which the concavity of the curve changes. We will differentiate the function twice to find the first and second derivative of the function. We will use the second derivative test to determine the concavity of the curve at different points.

Complete step by step answer:
Let us consider the given function $f\left( x \right)={{x}^{4}}-10{{x}^{3}}+24{{x}^{2}}+3x+5.$
First, we are going to find the first derivative of the equation with respect to $x.$
We will use the rules of differentiation to find the first derivative. The basic rule we should remember is $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}.$
Now we will get $\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( {{x}^{4}}-10{{x}^{3}}+24{{x}^{2}}+3x+5 \right).$
That is $\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}{{x}^{4}}-\dfrac{d}{dx}10{{x}^{3}}+\dfrac{d}{dx}24{{x}^{2}}+\dfrac{d}{dx}3x+\dfrac{d}{dx}5.$
The derivative of a constant is zero.
So, we will get ${f}'\left( x \right)=4{{x}^{3}}-3\times 10{{x}^{2}}+2\times 24x+3.$
That is, ${f}'\left( x \right)=4{{x}^{3}}-30{{x}^{2}}+48x+3.$
We will differentiate this again with respect to $x$ to find the second derivative of $f\left( x \right).$
We know that ${f}''\left( x \right)=\dfrac{{{d}^{2}}}{d{{x}^{2}}}f\left( x \right)=\dfrac{d}{dx}\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}{f}'\left( x \right).$
So, we will get ${f}''\left( x \right)=\dfrac{d}{dx}{f}'\left( x \right)=\dfrac{d}{dx}\left( 4{{x}^{3}}-30{{x}^{2}}+48x+3 \right).$
We will apply the rule $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ again.
So, this will give us ${f}''\left( x \right)=\dfrac{d}{dx}{f}'\left( x \right)=\dfrac{d}{dx}4{{x}^{3}}-\dfrac{d}{dx}30{{x}^{2}}+\dfrac{d}{dx}48x+\dfrac{d}{dx}3.$
Since the derivative of a constant is zero, we will get ${f}''\left( x \right)=3\times 4{{x}^{2}}-2\times 30x+48+0.$
That is, ${f}''\left( x \right)=12{{x}^{2}}-60x+48.$
Since $12$ is a common factor, let us take $12$ out to get ${f}''\left( x \right)=12\left( {{x}^{2}}-5x+4 \right).$
Consider the polynomial of degree $2$ in $x,$ ${{x}^{2}}-5x+4.$
Now, we know that $-1+-4=-5$ and $-1\times -4=4.$
So, we can write ${{x}^{2}}-5x+4=\left( x-1 \right)\left( x-4 \right).$
Thus, the second derivative of the function $f$ can be written as ${f}''\left( x \right)=12\left( x-1 \right)\left( x-4 \right).$
Therefore, the second derivative of the given function exists and ${f}''\left( x \right)=0$ at $x=1$ and $x=4.$
We will use the second derivative test for concavity to find the intervals in which the curve of the given function is concave up and the intervals in which the curve of the given function is concave down.
Let us consider the interval $\left( -\infty ,1 \right).$
In this interval, $x-1<0$ and $x-4<0.$ Therefore, ${f}''$ is positive. By the second derivative test, the curve of the function in this interval is concave up.
Let us consider the interval $\left( 1,4 \right).$
In this interval, $x-1>0$ and $x-4<0.$ Therefore ${f}''$ is negative. By the second derivative test, the curve of the function in this interval is concave down.
Let us consider the interval $\left( 4,\infty \right).$
In this interval, $x-1>0$ and $x-4>0$ implies ${f}''$ is positive. By the second derivative test, the curve of the function in this interval is concave up.
Therefore, the concavity of the curve changes at $x=1$ and $x=4.$
We will get $f\left( 1 \right)={{1}^{4}}-10\times {{1}^{3}}+24\times {{1}^{2}}+3\times 1+5=1-10+24+3+5=23.$
Also, $f\left( 4 \right)={{4}^{4}}-10\times {{4}^{3}}+24\times {{4}^{2}}+3\times 4+5=256-640+384+12+5=17.$
So, the points of inflection are $\left( 1,23 \right)$ and $\left( 4,17 \right).$

Note: A point where the graph of a function has a tangent line and where the concavity changes is called a point of inflection. At a point of inflection, the second derivative of $f$ either equals zero or does not exist.