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The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:
A$x = 0$
B$x = 1$ (max value $1/2$)
C$x = \sqrt 3$
D$x \to \infty$
Answer & Solution
Correct answer: B. $x = 1$ (max value $1/2$)
$f'(x) = ((1+x^2) - x(2x))/(1+x^2)^2 = (1 - x^2)/(1+x^2)^2 = 0$ ⇒ $x = 1$ (in [0,∞)). $f(1) = 1/2$. Check: $f(0) = 0$, $f \to 0$ as $x \to \infty$. So max = 1/2 at $x = 1$.
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