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.

Question

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!

Answer 1

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.

Answer 2

Hint: $x=\frac{1}{2}$ & $x=-2$ are the critical points. You need to check function in following intervals $(-\infty,-2),(-2,1/2),(1/2,+\infty)$

Answer 3

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 $\in (-2,\frac{1}{2})$?

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! 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) $$f(x)= \begin{cases} 2x-1-3(2x+4)+7~ \text{, x$\gt \frac{1}{2}$}\\ -(2x-1)+3(2x+4)+7~ \text{, x$\lt -2$}\\ -(2x-1)-3(2x+4)+7 ~\text{, x$\in(-2,\frac{1}{2})$} \end{cases} $$

Answer 4

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