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Question

The range of $f (x) = \frac{x^{4}}{1 + x^{8}}$ is
$[0, \infty)$
$[0, \frac{1}{2}]$
$[0, 1]$
$(- \infty, \infty)$

A

[0, \infty)

B

(- \infty, \infty)

C

[0, \frac{1}{2}]

D

[0, 1]

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Solution

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$\begin{matrix} W e h a v e f (x) = \frac{x^{4}}{1 + x^{8}} y = \frac{t}{1 + t^{2}} t \geq 0 y = \frac{1}{t + \frac{1}{t}} {\begin{matrix} f o r t = 0 y = 0 \end{matrix} y \leq \frac{1}{2} y \in [0, \frac{1}{2}] H e n c e, t h e o p t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

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