Question

@Nurali @whpalmer4 break the following function into intervals.

y = | x -1 | + | x - 2 |

Answer 1

That will give you two intervals. Same with \(|x-2|\). You just need to find where there is overlap.

Answer 2

Well, \(x-1\) is either negative or not: \[
x-1<0 \quad \quad 0\leq x-1
\]

Answer 3

What if the given function is given to be differentiate, what to do explain it briefly ....

Answer 4

you need to differentiate the function? use chain rule.

Answer 5

\[
|x| = \sqrt{x^2}
\]

Answer 6

How please Explain it stepwise and deeply

Answer 7

Okay, solve \(x-1<0\). Can you do it?

Answer 8

then solve \(x-2<0\).

Answer 9

Then you have \(x<a\), the intervals \(a\leq x<b\), and \(b\leq x\)

Answer 10

not getting it

Answer 11

\[
|x-a| = \begin{cases}-(x-a)&x<a\\ x-a&a\leq x\end{cases}
\]

Answer 12

\[
(f\circ g)(x) = |x|= \sqrt{x^2}\\
f(x) = \sqrt{x} \quad f'(x)=\frac{1}{2\sqrt x}\\
g(x) = x^2\quad g'(x) = 2x \\
(f\circ g)'(x) = (f'\circ g)(x) \cdot g'(x) = \frac{1}{2\sqrt{x^2}}(2x) = \frac{x}{\sqrt{x^2}}=\frac{x}{|x|}
\]

Answer 13

In short \[
|x|' = \frac{x}{|x|}
\]

Answer 14

Is that enough for you?

Answer 15

Yes Sir

Answer 16

@goformit100 I suggest you sit down and try to graph specific points on this graph yourself. See what it looks like. Sure it will take a little time, but you'll understand through self discovery.

Remember, the turning points of an absolute value are when it equals 0 to go between negative and positive.

Answer 17

Thanks for the suggestion Sir @Kainui