Answer:
=(a-b)(a^2+ab+b^2)
= a^3+a^2b+ab+ab^2-ba^2-ab^2-b^3
SIMPLIFY
=a^3+ab-b^3
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.
Answer:
Part A:
-Minimum: 10
-Q1: 17.5
-Median: 30
-Q3: 42.5
-Maximum: 50
Step-by-step explanation:
Part B: IQR= 25
This shows that the data varies for 25 different numbers. That HALF of the data is between 25 numbers.
Part C: Using a box-and-whisker plot you can interpret the different values. Minimum is the first dot (10), connected to the first line (Q1 which is 17.5), connected by a box to the median (30), connected by a box to the third line (Q3 which is 42.5), connected to the last dot which is the maximum (50). And IQR is Q3-Q1, so 42.5-17.5 which is 25.
Answer:
Do you have a picture of the problem so I can see it