Question

If $f\left(x\right)=\frac{x^2-1}{x^2+1}$, for every real $x$, then the minimum value of $f$

• A. does not exist because $f$ is unbounded
• B. is not attained even though $f$ is bounded
• C. is equal to $1$
• D. is equal to $-1$

Answer 1

The correct option is D

is equal to $-1$

Explanation for the correct option:

Step 1: Find the $f\text{'}\left(x\right)$ and critical point:

Given that,

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

Differentiate the above equation with respect to $x$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{\left(x^2+1\right){\displaystyle \frac{d\left(x^2-1\right)}{dx}}-\left(x^2-1\right){\displaystyle \frac{d\left(x^2+1\right)}{dx}}}{{\left(x^2+1\right)}^2}\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}\right]\\ & =& \frac{\left(x^2+1\right)\left(2x\right)-\left(x^2-1\right)\left(2x\right)}{{\left(x^2+1\right)}^2}\\ & =& \frac{2x\left[x^2+1-x^2+1\right]}{{\left(x^2+1\right)}^2}\\ & =& \frac{4x}{{\left(x^2+1\right)}^2}\end{array}$

For critical point of $x$ ,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 0\\ & \Rightarrow & x=0\end{array}$

Step 2: Find the $f\text{'}\text{'}\left(x\right)$ and check whether it is minimum or not:

$\begin{array}{rcl}f\text{'}\text{'}\left(x\right)& =& \frac{{\left(x^2+1\right)}^2\left(4\right)-4x\left(2\left(x^2+1\right)2x\right)}{{\left(x^2+1\right)}^4}\\ & =& \frac{\left(x^2+1\right)\left[4\left(x^2+1\right)-16x^2\right]}{{\left(x^2+1\right)}^4}\\ & =& \frac{-12x^2+4}{{\left(x^2+1\right)}^3}\end{array}$

At, $x=0$ ,

$f\text{'}\text{'}\left(0\right)>0$

Therefore, there is only one point of minima at $x=0$,

Step 3: Finding the minimum value of $f$:

Now, substitute $x=0$ in the given equation, we get the minimum value

$\begin{array}{rcl}f\left(0\right)& =& \frac{{\left(0\right)}^2-1}{{\left(0\right)}^2+1}\\ & =& -1\end{array}$

Hence, the correct option is D.