If f(x) = frac{x^2 - 1}{x^2 + 1} for every re | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = \frac{x^2 - 1}{x^2 + 1}$.We can rewrite the function as $f(x) = \frac{x^2 + 1 - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$.Since $x^2 \ge

Vedclass · Accepted Answer

Is equal to $-1$

Vedclass · Answer

(D) Given $f(x) = \frac{x^2 - 1}{x^2 + 1}$.We can rewrite the function as $f(x) = \frac{x^2 + 1 - 2}{x^2 + 1} = 1 - \frac{2}{x^2 + 1}$.Since $x^2 \ge 0$ for all real $x$,we have $x^2 + 1 \ge 1$,which implies $0 Subtracting this inequality from $1$,we get $1 - 2 \le 1 - \frac{2}{x^2 + 1} Thus,$-1 \le f(x) The minimum value is attained when $x^2 = 0$,i.e.,$x = 0$.$f(0) = \frac{0^2 - 1}{0^2 + 1} = -1$.Therefore,the minimum value of $f$ is $-1$.

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

Similar Questions

The domain of the function $f(x) = \log_e(x - [x])$ is

The domain of the function $f(x) = \frac{1}{\sqrt{|x|-x}}$ is

The domain of $f(x) = [\sin x] \cos \left( \frac{\pi}{[x - 1]} \right)$ is (where $[.]$ denotes the Greatest Integer Function $G.I.F.$).

If $f: R \rightarrow R$ is defined by $f(x)=[2x]-2[x]$ for $x \in R$,where $[x]$ is the greatest integer not exceeding $x$,then the range of $f$ is:

The domain of the function $f(x) = \log |\log x|$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module