Question

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

• A. Does not exist bacause $f$ 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: Simplify the given expression

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

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

Step 2: Solve for the minimum value

It is known that the square of a number is always non-negative.

So, $x^2\ge 0$

$$\begin{gathered} \Rightarrow x^2+1\ge 1 \\ \Rightarrow \frac{1}{x^2+1}\le 1 \\ \Rightarrow \frac{2}{x^2+1}\le 2 \\ \Rightarrow -\frac{2}{x^2+1}\ge -2 \\ \Rightarrow 1-\frac{2}{x^2+1}\ge 1-2 \\ \Rightarrow f\left(x\right)\ge -1 \end{gathered}$$

Thus, the minimum value of $f$ is equal to $-1$.

Hence, option(D) is the correct option.