The range of the function f(x) = x^2 + frac{1 | vedclass

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

Question

(A) Let $y = x^2 + \frac{1}{x^2+1}$.Let $t = x^2$. Since $x \in \mathbb{R}$,$t \geq 0$.Then $y = t + \frac{1}{t+1} = (t+1) + \frac{1}{t+1} - 1$.Since

Vedclass · Accepted Answer

$[1, \infty)$

Vedclass · Answer

(A) Let $y = x^2 + \frac{1}{x^2+1}$.Let $t = x^2$. Since $x \in \mathbb{R}$,$t \geq 0$.Then $y = t + \frac{1}{t+1} = (t+1) + \frac{1}{t+1} - 1$.Since $t \geq 0$,$t+1 \geq 1$.Using the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$ for $t+1$ and $\frac{1}{t+1}$:$\frac{(t+1) + \frac{1}{t+1}}{2} \geq \sqrt{(t+1) \cdot \frac{1}{t+1}} = 1$.So,$(t+1) + \frac{1}{t+1} \geq 2$.Therefore,$y = (t+1) + \frac{1}{t+1} - 1 \geq 2 - 1 = 1$.The minimum value is attained at $t=0$ (i.e.,$x=0$),where $f(0) = 0 + \frac{1}{0+1} = 1$.Thus,the range of the function is $[1, \infty)$.

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

Similar Questions

Function $f:(2, \infty) \rightarrow R$ defined by $f(x) = x^2 - 4x + 5$. The range of $f$ is $=$ . . . . . . .

The set of all real values of $x$ such that $f(x) = \sqrt{\frac{[x]-1}{[x]^2-[x]-6}}$ is a real-valued function is

Let $f = \left\{ \left(x, \frac{x^2}{1+x^2} \right) : x \in R \right\}$ be a function from $R$ into $R$. Determine the range of $f$.

If $[x]$ represents the greatest integer $\leq x$, then the range of the real-valued function $f(x) = \frac{1}{\sqrt{[x]^2+[x]-2}}$ is

The domain of the function $f: R \rightarrow R$ defined by $f(x) = \sqrt{x^{2}-7x+12}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module