Range of the function f(x) = frac{x^2}{x^2 + | vedclass

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

Question

(C) Let $y = \frac{x^2}{x^2 + 1}$.Since $x^2 \ge 0$ for all real $x$,the numerator is always non-negative and the denominator is always greater than t

Vedclass · Accepted Answer

$[0, 1)$

Vedclass · Answer

(C) Let $y = \frac{x^2}{x^2 + 1}$.Since $x^2 \ge 0$ for all real $x$,the numerator is always non-negative and the denominator is always greater than the numerator.As $x 	o \pm \infty$,$y 	o 1$,but $y$ never reaches $1$ because $x^2 At $x = 0$,$y = \frac{0}{0+1} = 0$.Since $x^2 \ge 0$ and $x^2 + 1 > 0$,the value of $y$ is always non-negative.Thus,the range of the function is $[0, 1)$.

Range of the function $f(x) = \frac{x^2}{x^2 + 1}$ is

Similar Questions

Domain of the function $f(x) = \frac{x^2 - 3x + 2}{x^2 + x - 6}$ is

The domain of the function $f(x) = [\log_{10}(\frac{5x - x^2}{4})]^{1/2}$ is

The domain of the real valued function $f(x) = \frac{\sqrt{6x^2+5x-6}}{\sqrt{4-x}-\sqrt{x+4}}$ is

The domain of the function $f(x) = \exp (\sqrt {5x - 3 - 2{x^2}} )$ is

The range of the real valued function $f(x) = \frac{x^2+x+1}{x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module