Range of x^2/x^4 + 1.

Question

Range of \(\cfrac{x^2}{x^4+1}\).

Answer 1

f(x) = \(\cfrac{x^2}{x^4+1}\) = \(\cfrac{1}{x^2+\cfrac1{x^2}}\)

= \(\cfrac{1}{\left(x+\cfrac1{x}\right)^2-2}\)

\(\because\) \({\left(x+\cfrac1{x}\right)^2}\) \(\geq\) 2² or \({\left(x+\cfrac1{x}\right)^2}\) \(\geq\) 4

= \({\left(x+\cfrac1{x}\right)^2}\) - 2 \(\geq\) 4 - 2

= \({\left(x+\cfrac1{x}\right)^2}\) - 2 \(\geq\) 2

= \(\cfrac{1}{\left(x+\cfrac{1}{x}\right)^2-2}\)  \(\leq\) \(\cfrac12\)

= \(\cfrac{x^2}{x^4+1}\) \(\leq\) \(\cfrac12\)....(i)

Also, x² \(\geq\) 0 and x⁴ + 1 \(\geq\) 1

= \(\cfrac{x^2}{x^4+1}\) \(\geq\) 0 ...(ii)

(\(\because\) Division of two positive numbers be always positive)

Then from (i) and (ii),we obtain

0 \(\leq\) \(\cfrac{x^2}{x^4+1}\) \(\leq\) \(\cfrac12\)

\(\therefore\) Range of f(x) = \(\cfrac{x^2}{x^4+1}\) is [0,\(\cfrac12\)].