Solve $|x| + |x-1| <5$

Question

Solve $|x| +| x-1| <5$

Attempt:

We have $4$ cases:

$1.\quad x>0$ and $x>1\implies x>1$

$2.\quad x>0$ and $x<1$

$3.\quad x<0$ and $x<1 \implies x<0$

$4.\quad x<0$ and $x>1$ can't happen

For the first case:

$x+x-1<5\\2x<6\\ x<3$

For the second case:

if $x<0$ so $x<3$ true for all $x$

For the third case:

this is also true for all $0<x<1$

My final answer is wrong. Can someone show me how to solve this?

Answer 1

If you want to solve it by considering of cases, so we have three cases:

1. $x>1$, which gives $2x-1<5$ and $1<x<3$;

2. $0<x\leq1$, which gives $x-x+1<5$ and $0<x\leq1$;

3. $x\leq0$, which gives $-x-x+1<5$ and $-2<x\leq0$.

Finely we get $-2<x<3$.

Answer 2

Since $f(x)=|x|+|x-1|$ is a convex function, the equation $$|x|+|x-1|=5$$ has two roots maximum.

But $3$ and $-2$ are roots, which gives the answer $-2<x<3$.

Answer 3

Case 1 : Assuming $x \geq 1$ :

• $(x) + (x-1) < 5 \implies 2x < 6 \implies x < 3$. Therefore $1 \leq x < 3$ is a solution.

Case 2 : Assuming $0 \leq x < 1 $ :

• $(x) + (-x + 1) < 5 \implies 1 < 5$. Which is always true, so $0 \leq x < 1$ is valid.

Case 3 : Assuming $x <0$ :

• $(-x) + (-x + 1) < 5 \implies -2x < 4 \implies x > -2$, so $-2 < x < 0$ is a solution.

Overall the inequality is true in $]-2,3[$.

Answer 4

Hint: think of it geometrically, with $x$ being a point on the real axis.

Solve $\mid x\mid +\mid x-1\mid <5$

$|x|$ is the distance to the origin $0\,$, and $|x-1|$ is the distance to point $1$.

If $x$ is outside segment $[0,1]$ then the sum of the two distances is twice the distance to the midpoint, so the inequality reduces to $2\mid x-\frac{1}{2}\mid \lt 5 \iff -\frac{5}{2} \;\lt\; x - \frac{1}{2} \lt \frac{5}{2} \;\iff \; -2 \lt x \lt 3\,$.

If $x \in [0,1]$ then the sum of the two distances is the length of the segment $\mid x\mid +\mid x-1\mid = 1 \lt 5\,$, so in the end the solution set is $x \in [-2,3]$.

Answer 5

I'll add my tuppence for what it is worth. Since both sides of the relation are positive, we can square without changing the inequality:

$$\begin{align} ∣x∣+∣x−1∣\,<\,5 \quad&\Longrightarrow\quad (∣x∣+∣x−1∣)^{2}\,<\,25 \\ \quad&\Longrightarrow\quad |x|^{2} + 2|x|\,|x-1| + |x-1|^{2} \,<\,25 \\ \quad&\Longrightarrow\quad 2x^{2} - 2x + 1 + 2|x(x-1)| \,<\,25 \\ \quad&\Longrightarrow\quad |x^{2}-x| \,<\,x - x^{2} + 12, \end{align}$$

since $|a|^{2}=a^{2}$ for any $a\in\mathbb{R}$ and $|a|\cdot|b|=|ab|$. Now, since $|x|<M$ implies $-M<x<-M$, this inequality reduces to $$\begin{align} |x^{2}-x| \,<\,x - x^{2} + 12 \quad&\Longrightarrow\quad x^{2} - x - 12 \,<\,x^{2}-x \,<\,x - x^{2} + 12 \\ \quad&\Longrightarrow\quad 0 \,<\,12 \,<\,2x - 2x^{2} + 24 \\ \quad&\Longrightarrow\quad 0 \,<\,6 \,<\,x - x^{2} + 12 \end{align}$$

Thus, we clear need to solve $6<x-x^{2}+12$:

$$\begin{align} 6 \,<\,x - x^{2} + 12 \quad\Longrightarrow\quad x^{2}-x-6 \,=\, (x+2)(x-3)\,<\,0 \end{align}$$

which has solutions for $-2<x<3$.