Q. 1 - Calculus

Question

Sign analyses for derivatives: For each function f that follows, find the derivative $f\text{'}.$ Then determine the intervals on which the derivative $f\text{'}$ is positive and the intervals on which the derivative $f\text{'}$ is negative. Record your answers on a sign chart for $f\text{'},$ with tick-marks only at the x-values where $f\text{'}$  is zero or undefined.

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

Step-by-Step Solution

Answer

The derivative of the given function is positive on the interval $\left(-1,1\right)$ and negative on the intervals $\left(-\infty ,-1\right) \mathrm{and} \left(1,\infty \right).$ The derivative of the function is zero at $x=1 \mathrm{and} x=-1.$

Step 1. Given Information.

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

Step 2. Find the derivative.

To find the derivative of the given function, we will use the quotient rule of differentiation.

So,

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

Now, let's find the critical point by putting the above equation to zero,

$\begin{array}{c}\frac{-x^2+1}{{\left(x^2+1\right)}^2}=0\\ -x^2+1=0\\ x^2=1\\ x=\pm 1\end{array}$

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

Step 3. Determine the intervals.

The intervals we get by the critical points are $(-\mathrm{\infty },-1),(-1,1)\text{ and }(1,\mathrm{\infty }).$

Now, let's take the interval $(-\mathrm{\infty },-1),$to determine where the derivative of the function is positive or negative.

Let $x=-2$

$$\begin{gathered} f\text{'}(-2)=\frac{-{(-2)}^2+1}{{\left((-2{)}^2+1\right)}^2} \\ f\text{'}(-2)=\frac{-4+1}{{\left(5\right)}^2} \\ f\text{'}(-2)=\frac{-3}{25} \\ f\text{'}(-2)=-0.12 \end{gathered}$$

Since $-0.12<0,$ thus the $f\text{'} \mathrm{is} \mathrm{negative} \mathrm{on} \mathrm{the} \mathrm{interval} \left(-\infty ,-1\right).$

Now, the interval $\left(-1,1\right)$

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

Since $1>0,$ thus the $f\text{'} \mathrm{is} \mathrm{positive} \mathrm{on} \mathrm{the} \mathrm{interval} \left(-1,1\right).$

Step 4. Determine the intervals.

Now, the interval $\left(1,\infty \right),$

$$\begin{gathered} \mathrm{Let} x=2 \\ f\text{'}\left(0\right)=\frac{-(2{)}^2+1}{{\left((2{)}^2+1\right)}^2} \\ f\text{'}\left(0\right)=\frac{-3}{25} \\ f\text{'}\left(0\right)=-0.12 \end{gathered}$$

Since $-0.12<0$ thus the $f\text{'} \mathrm{is} \mathrm{negative} \mathrm{on} \mathrm{the} \mathrm{interval} \left(1,\infty \right).$