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Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value.

I thought of it this way: $$f(x)=\begin{cases}2x-1-3(2x+4)+7 \,(\text{then I simplify)} & \text{if $x>0$}\\ -(2x-1-3(2x+4)+7)\,(\text{then I simplify)} & \text{if $x\le0$}\end{cases}$$

But is there some way without having to use the cases?

Edit: NEW work on this problem! I found three cases;

If $x\in ]-\infty,-2]$ then f(x)=$4x+20$

If $x\in]-2,1/2]$ then f(x)=$-8x-4$

If $x\in]1/2,+\infty[$ then f(x)=$-4x-6$

IS THIS TRUE?

Thank you very much!

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    $\begingroup$ This is not correct. When is 2x-1 positive or negative? And the same question about 2x+4. It has nothing to do with $x>0$ and $x<0$. $\endgroup$
    – Poppy
    Dec 22, 2013 at 12:10
  • $\begingroup$ So how can I do it?? $\endgroup$
    – Bambaraku
    Dec 22, 2013 at 12:12
  • $\begingroup$ Imagine you have only $f(x)=|2x-1|$. Try graphing it online or even better by yourself. The critical point is $\frac{1}{2}$. $\endgroup$
    – Poppy
    Dec 22, 2013 at 12:15
  • $\begingroup$ Each of the absolute value expressions has different regions where you can 'simplify' them. You need to find all of them (In general, in an expression with 2 abs. values, there will be 4 regions: both positive, first pos. second neg, first neg. second pos, both negative.) $\endgroup$ Dec 22, 2013 at 12:15
  • $\begingroup$ So even if I do that I'll have to make a function defined by cases ? $\endgroup$
    – Bambaraku
    Dec 22, 2013 at 12:16

4 Answers 4

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Here is a start. You need to consider different cases.

Case 1: $2x-1\geq 0 \cap 2x+4 \geq 0 \implies x\geq \frac{1}{2} $. For this case, we have

$$ f(x) = (2x-1)-3(2x+4)+7=-4x -6 .$$

Case 2: $2x-1 < 0 \cap 2x+4 < 0 \implies x<-2$ which gives

$$ f(x) = -(2x-1)-3(-(2x+4))+7=4x + 20. $$

Now, I leave it for you to discover the other possible cases.

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  • $\begingroup$ Please check my answer: math.stackexchange.com/a/615628/117228 $\endgroup$
    – Bambaraku
    Dec 22, 2013 at 15:17
  • $\begingroup$ Yes, it is correct. $\endgroup$ Dec 22, 2013 at 16:19
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    $\begingroup$ THANK YOU SO MUCH!! Then how can I use it to solve the equation $f(x)=0$ $\endgroup$
    – Bambaraku
    Dec 22, 2013 at 16:22
  • $\begingroup$ Make sure of the interval in your second case . It should be open at $1/2$ and closed at $-2$. $\endgroup$ Dec 22, 2013 at 16:28
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    $\begingroup$ For the third case it should be $[1/2, \infty) $. $\endgroup$ Dec 22, 2013 at 16:40
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Hint: $x=\frac{1}{2}$ & $x=-2$ are the critical points. You need to check function in following intervals $(-\infty,-2),(-2,1/2),(1/2,+\infty)$

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  • $\begingroup$ Do you mean $]-\infty,-2]$ or $]-\infty,-2[$ $\endgroup$
    – Bambaraku
    Dec 22, 2013 at 14:44
  • $\begingroup$ $(-\infty,-2)$ because for $x=-2,f(-2)=|2(-2)-1|+3|2(-2)+4|+7|=12$ $\endgroup$
    – Adi Dani
    Dec 22, 2013 at 15:04
  • $\begingroup$ Adil, please see my answer here: math.stackexchange.com/a/615628/117228 $\endgroup$
    – Bambaraku
    Dec 22, 2013 at 15:05
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Let f(x)=|x|.We know $$f(x)=\begin{cases} -x \text{ if x<0}\\ x\text{ if x>0} \end{cases}$$

What if f(x)=|x-1|. We must have, $$f(x)=\begin{cases} -(x-1) \text{ if x<1}\\ x-1\text{ if x>1} \end{cases}$$

Note that there are 3 terms in the given function. $|2x-1|$, $3|2x+4|$ and 7.

The first term is postive if x$\gt \frac{1}{2}$. The second term is positive if $x\gt -2$. Hence we must have, $$f(x)=2x-1-3(2x+4)+7 \text{ whenever x$\gt \frac{1}{2}$}$$ The first term is negative if x$\lt \frac{1}{2}$ and the second term is negative if x$\lt -2$.Hence we have a second condition, $$f(x)=-(2x-1)+3(2x+4)+7 \text{ whenever x$\lt -2$}$$ Can you work out what will happen if x $\in (-2,\frac{1}{2})$?


! If x $\in (-2,\frac{1}{2})$ surely |2x-1|<0,hence |2x-1|=-(2x-1) and |2x+4|>0,hence |2x+4|=2x+4. Finally we have the following for f(x) $$f(x)= \begin{cases} 2x-1-3(2x+4)+7~ \text{, x$\gt \frac{1}{2}$}\\ -(2x-1)+3(2x+4)+7~ \text{, x$\lt -2$}\\ -(2x-1)-3(2x+4)+7 ~\text{, x$\in(-2,\frac{1}{2})$} \end{cases} $$

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  • $\begingroup$ No :( I don't know $\endgroup$
    – Bambaraku
    Dec 22, 2013 at 14:40
  • $\begingroup$ Did you understand this: if f(x)=|x|.We know $$f(x)=\begin{cases} -x \text{ if x<0}\\ x\text{ if x>0} \end{cases}$$ $\endgroup$
    – GTX OC
    Dec 22, 2013 at 15:00
  • $\begingroup$ Please check my answer: math.stackexchange.com/a/615628/117228 $\endgroup$
    – Bambaraku
    Dec 22, 2013 at 15:03
  • $\begingroup$ So is it: I found three cases; <br> If $x\in ]-\infty,-2]$ then f(x)=$4x+20$ <br> If $x\in]-2,1/2]$ then f(x)=$-8x-4$ <br> If $x\in]1/2,+\infty[$ then f(x)=$-4x-6$ <br> IS THIS TRUE? $\endgroup$
    – Bambaraku
    Dec 22, 2013 at 16:08
  • $\begingroup$ Yes. It is correct. $\endgroup$
    – GTX OC
    Dec 23, 2013 at 1:52
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The absolute value has a discontinuous slope change at zero. Hence, f(x) has discontinuous slope changes at -2 from $|2x+4|$ and at 1/2 from $|2x-1|$. So there are three cases: $-\infty < x \le -2$, $-2 < x \le 1/2 $ and $1/2 < x < \infty$.

Now for $x$ sufficiently negative $|2x+4| = -2x-4$ and we know that $|2x+4|$ had discontinuous slope at $2x+4=0$. Similarly for sufficiently negative $x$ $|2x-1| = -2x+1$ and changes slope when $2x-1=0$. Hence

$$ f(x) = \left{ \begin{array}{ll} +3(2x+4)-(2x-1)+7, & x \le -2; \\ -3(2x+4) -(2x-1)+7, & -2 <x \le 1/2; \\ -3(2x+4)+(2x-1)+7, & 1/2 < x \right. $$

Sorry! Don't see why the formatting is messed up. Hope it makes sense! I tried to use plain vanilla latex commands. I need to learn MathJaX

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