What is the absolute extrema of the function: #2x/(x^2 +1)# on closed interval [-2,2]?

Question

Emma Aldridge · Accepted Answer

The absolute extrema of a function in a closed interval #[a,b]# can be or local extrema in that interval, or the points whose ascissae are #a or b#.

So, let's find the local extrema:

#y'=2* (1*(x^2+1)-x*2x)/(x^2+1)^2=2*(-x^2+1)/(x^2+1)^2#.

#y'>=0#

if

#-x^2+1>=0rArrx^2<=1rArr-1<=x<=1#.

So our function is decresing in #[-2,-1)# and in #(1,2]# and it is growing in #(-1,1)#, and so the point #A(-1-1)# is a local minimum and the point #B(1,1)# is a local maximum.

Now let's find the ordinate of the points at the extrema of the interval:

#y(-2)=-4/5rArrC(-2,-4/5)#

#y(2)=4/5rArrD(2,4/5)#.

So the candidates are:

#A(-1-1)# #B(1,1)# #C(-2,-4/5)# #D(2,4/5)#

and it's easy to understand that the absolute extrema are #A# and #B#, as you can see:

graph{2x/(x^2 +1) [-2, 2, -5, 5]}

Charles Anctil · Answer

undefined To find the absolute extrema of the function $ f(x) = \frac{2x}{x^2 + 1} $ on the closed interval $[-2, 2]$, we follow these steps:

1. Find the critical points of the function within the interval by finding where the derivative equals zero or is undefined.
2. Evaluate the function at the critical points and the endpoints of the interval.
3. The highest and lowest values obtained from these evaluations will be the absolute maximum and minimum, respectively.

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

Critical points:
$$ f'(x) = 0 $$
$$ \frac{2}{(x^2 + 1)^2} = 0 $$
$$ 2 = 0 $$
$$ \text{No critical points} $$

The derivative is never undefined, so there are no critical points within the interval.

Now, we evaluate the function at the endpoints and the critical points (which we found none):
$$ f(-2) = \frac{2(-2)}{(-2)^2 + 1} = \frac{-4}{5} $$
$$ f(2) = \frac{2(2)}{(2)^2 + 1} = \frac{4}{5} $$

Since there are no critical points within the interval, the only potential extrema occur at the endpoints.

Therefore, the absolute extrema of $ f(x) $ on the interval $[-2, 2]$ are:
Absolute maximum: $ f(2) = \frac{4}{5} $
Absolute minimum: $ f(-2) = \frac{-4}{5} $