Answer:
66.8421% decrease
Step-by-step explanation:
See photos for solutions and steps :)
Substitution for both equations is (-3,1)
By definition of absolute value, you have

or more simply,

On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For <em>x</em> > -1, we have
(<em>x</em> + 1)<em>'</em> = 1
while for <em>x</em> < -1,
(-<em>x</em> - 1)<em>'</em> = -1
More concisely,

Note the strict inequalities in the definition of <em>f '(x)</em>.
In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:


All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.
7x-32=3
7x=3+32
7x=35
x=35/7
x=5
HOPE IT HELPS U!!
have a great day ahead ! :)