How do you find the intervals of increasing and decreasing using the first derivative given #y=1/(x+1)^2#?

Question

Isaiah Allen · Accepted Answer

#y# is increasing for #(-oo, -1)# and decreasing for #(-1, +oo)#

#y = 1/(x+1)^2#

#= (x+1)^-2#

Applying the Power Rule and Chain Rule

#y' = -2(x+1)^-3 * 1 #

#y'# represents the slope of #y# at an point #x# in its domain

Hence: #y# will be increasing when #y' >0# and decreasing when #y'<0#

#:. y # will be increasing for #-2(x+1)^-3 >0# #-> (x+1)<0# i.e. #x< -1#

and #y # will be decreasing for #-2(x+1)^-3 <0# #-> (x+1)>0# i.e. #x> -1#

Hence: #y# is increasing for #(-oo, -1)# and decreasing for #(-1, +oo)#

This can be seen from the graph of #y# below:

graph{1/(x+1)^2 [-3.847, 1.626, -0.537, 2.202]}

Aurora Alvarado · Answer

undefined To find the intervals of increasing and decreasing for the function $ y = \frac{1}{(x+1)^2} $ using the first derivative:

1. Find the first derivative of the function $ y $ with respect to $ x $.
2. Set the first derivative equal to zero and solve for $ x $ to find critical points.
3. Determine the intervals where the first derivative is positive or negative to identify increasing and decreasing intervals.

Now, let's proceed with the steps:

1. The first derivative of $ y $ with respect to $ x $ can be found using the chain rule:

$$ \frac{d}{dx}\left(\frac{1}{(x+1)^2}\right) = -\frac{2}{(x+1)^3} $$

2. Set the first derivative equal to zero and solve for $ x $ to find critical points:

$$ -\frac{2}{(x+1)^3} = 0 $$
$$ (x+1)^3 \neq 0 $$

The expression $ (x+1)^3 $ is never zero, so there are no critical points.

3. Determine the intervals where the first derivative is positive or negative:

The first derivative $ -\frac{2}{(x+1)^3} $ is negative for all $ x $ values except $ x = -1 $. At $ x = -1 $, the first derivative is undefined. Therefore, the function is decreasing for all $ x $ except at $ x = -1 $, where it's undefined.

So, the interval of decreasing is $ (-\infty, -1) \cup (-1, \infty) $. There are no intervals of increasing for this function.