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

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(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

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$[1, \infty)$

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(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

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