What are the absolute extrema of #f(x)=x/(x^2+1) in(0,2)#?

Question

Emily Ammerman · Accepted Answer

The first derivative is #df(x)/dx=(x'*(x^2+1)-x*(x^2+1)')/(x^2+1)^2=(1-x^2)/(1+x^2)^2# Hence the first derivative nullifies at #x=+-1# but because #x ε (0,2)# We have that at #x=1# we got a maximum which is #f(1)=1/2#

Axel Alvarez · Answer

undefined To find the absolute extrema of $ f(x) = \frac{x}{x^2 + 1} $ on the interval (0, 2), we first find the critical points by setting the derivative equal to zero and finding where it's undefined. Then, we evaluate $ f(x) $ at these critical points and at the endpoints of the interval (0 and 2). The largest and smallest values among these will be the absolute extrema.

1. Find critical points by finding where the derivative is zero or undefined:

$$ f'(x) = \frac{(x^2 + 1) - x(2x)}{(x^2 + 1)^2} = 0 $$

$$ \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = 0 $$

$$ x^2 + 1 - 2x^2 = 0 $$

$$ -x^2 + 1 = 0 $$

$$ x^2 = 1 $$

$$ x = \pm 1 $$

2. Evaluate $ f(x) $ at critical points and endpoints:

$$ f(0) = 0 $$
$$ f(1) = \frac{1}{2} $$
$$ f(2) = \frac{2}{5} $$

3. Compare the values:

$$ f(0) = 0 $$
$$ f(1) = \frac{1}{2} $$
$$ f(2) = \frac{2}{5} $$

Thus, the absolute maximum is $ \frac{1}{2} $ at $ x = 1 $, and the absolute minimum is $ 0 $ at $ x = 0 $.