Q. 5 - Calculus

Question

Sign analyses for second derivatives: Repeat the instructions of the previous block of problems, except find sign intervals for the second derivative $f\text{'}\text{'}$ instead of the first derivative.

$f\left(x\right)=\frac{x}{x^2+1}$

Step-by-Step Solution

Answer

The second derivative of the given function is the local maximum at $x=1,$ and a local minimum at $x=-1.$

Step 1. Given Information.

The given function is $f\left(x\right)=\frac{x}{x^2+1}.$

Step 2. Find the second derivative.

To find the second derivative, first, we find the first derivative then the second.

So,

$$\begin{gathered} f\left(x\right)=\frac{x}{\left(x^2+1\right)} \\ f\text{'}\left(x\right)=\frac{\left(x^2+1\right)-2x^2}{{\left(x^2+1\right)}^2} \\ f\text{'}\left(x\right)=\frac{1}{\left(x^2+1\right)}-\frac{2x^2}{{\left(x^2+1\right)}^2} \\ f\text{'}\text{'}\left(x\right)=\frac{-2x}{{\left(x^2+1\right)}^2}-\frac{{\left(x^2+1\right)}^{2}4x-4x^2\left(x^2+1\right)\cdot 2x}{{\left(x^2+1\right)}^4} \\ f\text{'}\text{'}\left(x\right)=\frac{-2x}{{\left(x^2+1\right)}^2}-\frac{4x\left(x^2+1\right)\left[x^2+1-2x^2\right]}{{\left(x^2+1\right)}^4} \\ f\text{'}\text{'}\left(x\right)=\frac{-2x}{{\left(x^2+1\right)}^2}-4x\frac{\left[-x^2+1\right]}{{\left(x^2+1\right)}^3} \\ f\text{'}\text{'}\left(x\right)=\frac{1}{{\left(x^2+1\right)}^3}\left[-2x^3-2x+4x^3-4x\right] \\ f\text{'}\text{'}\left(x\right)=\frac{1}{{\left(x^2+1\right)}^3}\left[-6x+2x^3\right] \\ f\text{'}\text{'}\left(x\right)=\frac{-2x\left(3+x^2\right)}{{\left(x^2+1\right)}^3} \end{gathered}$$

Step 3. Find sign intervals for the second derivative.

To find the sign intervals let's find the critical points from the first derivative.

So,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\left(x^2+1\right)-2x^2}{{\left(x^2+1\right)}^2} \\ 0=\frac{\left(x^2+1\right)-2x^2}{{\left(x^2+1\right)}^2} \\ 0=\left(x^2+1\right)-2x^2 \\ 0=-x^2+1 \\ x=\pm 1 \end{gathered}$$

Thus, the critical points are $x=1 \mathrm{and} x=-1.$

Now, at $x=1, f\text{'}\text{'}\left(x\right)<0$ and at $x=-1, f\text{'}\text{'}\left(x\right)>0.$