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Question I am referring to: enter image description here

Why does the absolute value portion of the expression of the function this graph corresponds to has to be 0 when x=1?

Here are the five choices: enter image description here

most venerable sir
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  • I think you must have omitted some context. I am guessing that you are plotting something like $x \mapsto {1 \over 2} (x-1 + |x-1|)$, and hence it is saying that you must have $|x-1| = 0$ at $x=1$. Why, I have no idea. – copper.hat Oct 25 '15 at 17:26
  • @copper.hat, there are five choices. I think they don't matter. – most venerable sir Oct 25 '15 at 17:32
  • I think the 'major change' is misleading. Some detective work is needed. I would look at $x=0$ which eliminates (A),(E), then $x=-1$ which eliminates (C) and then $x=2$ which eliminates (D). Then use the Conan Doyle principle. – copper.hat Oct 25 '15 at 17:37
  • Well, (B) is the only guy non-differentiable at $x=1$. – Christoph Oct 25 '15 at 17:38
  • As an aside, many things are 'obvious' when you know them. – copper.hat Oct 25 '15 at 17:38
  • @Christoph's answer is more sophisticated. – copper.hat Oct 25 '15 at 17:39
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    He know the answer but he is asking why the book said major changes in absolute values happened at the zeros of the function which include an absolute value right ? – IrbidMath Oct 25 '15 at 17:40
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    @Ameryr, yes I figured out the answer by plugging in. But the book throws me off by saying out of no where that since the "major change" happen at (1,1), the expression in the absolute value signs should be equal to 0. – most venerable sir Oct 25 '15 at 17:42
  • If we have a function like this $f(x) = |g(x)| + h(x)$ does this function has a major change at the zeros of $g(x)$ ? This is the question – IrbidMath Oct 25 '15 at 17:44
  • Yes, it has a major change when $g(x) = 0$ as long as $g(x) \neq 0$ close to the zeros of $g(x)$. Please consider the answer I posted 10 minutes ago. – eyqs Oct 25 '15 at 17:46
  • I think there should be conditions on $g,h$ continuous and differentiable. Or if they just linear – IrbidMath Oct 25 '15 at 17:47
  • If $g$ and $h$ were not differentiable then there would be major changes at every point where they are not differentiable. – eyqs Oct 25 '15 at 17:50

1 Answers1

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Answer B is correct. You know that the function is a sum of two different functions, a linear function and an absolute value function.

The absolute value function $f(x) = |g(x)|$ has a major change when $f(x) = 0$. But in this graph, there is a major change at $x = 1$, so $f(x) = 0$ at $x = 1$, and the absolute value function must be translated by one unit to the right to make the function work. In other words, $g(x) = x - 1$ (translated one unit to the right), and the answer with the $f(x) = |x-1|$ term is correct.

eyqs
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  • How do you see from jus the graph that it's sum of a linear and absolute value? – most venerable sir Oct 25 '15 at 17:36
  • Yep, because the major change you see at $(1,1)$ is characteristic of an absolute value function. If the slope of the absolute value function is the same as the slope of a linear function, then you'll end up with a constant function on one side, and a function with double the slope on the other side. – eyqs Oct 25 '15 at 17:37
  • Is what u stated a theorem? How adding one absolute to a linear will result in doubling the slope of the absolute one. – most venerable sir Oct 25 '15 at 17:39
  • Sorry, you keep editing your comments so I'll reply to your edited question. Consider the function $f(x) = x + |x|$. When $x < 0$, $f(x) = 0$, and when $x > 0$, $f(x) = 2x$. – eyqs Oct 25 '15 at 17:40
  • More importantly, all your answer choices only involve absolute value and linear functions! – eyqs Oct 25 '15 at 17:41