The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on …

Question

The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:

Accepted Answer

B. $x = 1$ (max value $1/2$)

We have $f'(x) = ((1+x^2) - x(2x))/(1+x^2)^2$.
That is $f'(x) = (1 - x^2)/(1+x^2)^2$.
So $f'(x) = 0$.
⇒ $x = 1$ (in [0,∞)). $f(1) = 1/2$.
**Check:** $f(0) = 0$, $f 	o 0$ as $x 	o \infty$.
We have **So max = 1/2 at.
So $x = 1$.**.

Answer

$x = 0$

Answer

$x = \sqrt 3$

Answer

$x 	o \infty$

The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:

Keep practising Unit 4-5: Applications of Derivatives & Curve Sketching