How do use the first derivative test to determine the local extrema #f(x) = x / (x^2+1)#?

Question

Cameron Alexander · Accepted Answer

See the explanation.

#f'(x)=(x^2+1-2x^2)/(x^2+1)^2=(1-x^2)/(x^2+1)^2# #f'(x)=0 <=> 1-x^2=0 <=> x=1 vv x=-1# #f'(x)>0# #AAx in (-1,1)# function increasing #f'(x)<0# #AAx in (-oo,-1)uu(1,oo)# function decreasing #f'(x)# changes sign at #x=-1# and #f(x)# has a minimum value #f_min=f(-1)=-1/2# #f'(x)# changes sign at #x=1# and #f(x)# has a maximum value #f_max=f(1)=1/2#

Charles Anctil · Answer

undefined To use the first derivative test to determine the local extrema of $ f(x) = \frac{x}{x^2 + 1} $, follow these steps:

1. Find the first derivative of the function, $ f'(x) $.
2. Set $ f'(x) $ equal to zero and solve for $ x $. These points are potential critical points where the function may have local extrema.
3. Determine the sign of $ f'(x) $ in the intervals defined by the critical points found in step 2.
4. If the sign of $ f'(x) $ changes from positive to negative at a critical point, then that point is a local maximum. If the sign changes from negative to positive, then it is a local minimum.
5. If $ f'(x) $ does not change sign at a critical point, then the test is inconclusive at that point.

Now, applying these steps to the function $ f(x) = \frac{x}{x^2 + 1} $:

1. Find $ f'(x) $ by using the quotient rule:
$$ f'(x) = \frac{(1)(x^2 + 1) - (2x)(x)}{(x^2 + 1)^2} $$
$$ f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} $$
$$ f'(x) = \frac{1 - x^2}{(x^2 + 1)^2} $$

2. Set $ f'(x) = 0 $ and solve for $ x $:
$$ 1 - x^2 = 0 $$
$$ x^2 = 1 $$
$$ x = \pm 1 $$

3. We have two critical points, $ x = -1 $ and $ x = 1 $. Now, determine the sign of $ f'(x) $ in the intervals $ (-\infty, -1) $, $ (-1, 1) $, and $ (1, \infty) $:

- For $ x < -1 $, $ f'(x) $ is positive.
   - For $ -1 < x < 1 $, $ f'(x) $ is negative.
   - For $ x > 1 $, $ f'(x) $ is positive.

4. Since $ f'(x) $ changes from positive to negative at $ x = -1 $ and from negative to positive at $ x = 1 $, we have a local maximum at $ x = -1 $ and a local minimum at $ x = 1 $.

Therefore, the local maximum occurs at $ x = -1 $ and the local minimum occurs at $ x = 1 $.