Split at the absolute-value boundary

x|x-2|\ge1

Total marks: 8

Solution

(a)

Absolute-value inequalities

Solve $x|x-2|\ge1$ exactly.

Explore this idea

Hint

Split at $x=2$ : for $x\ge2$ , $|x-2|=x-2$ ; for $x<2$ , $|x-2|=2-x$ .
Solve each case and intersect with its range of $x$ .
Check with a graph of $y=x|x-2|$ against $y=1$ .

Answer · (a)

x=1

Explore the idea

y = x|x − 2| against y = 1

The peak of the small hump between 0 and 2 just touches the line, giving one isolated solution.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

Y = x|x − 2|
Y = 1
Touching and crossing points

Ideas to try

Why is there no solution between

x=1

and

x=1+\sqrt2

(b)

Shifting a known result

Hence solve $(x+1)|x-1|\ge1$ .

Explore this idea

Hint

Compare with (a): replace $x$ by $x+1$ in $x|x-2|$ .
Then $x+1\in\{1\}\cup[1+\sqrt2,\infty)$ .
Subtract 1 to get $x$ .

Answer · (b)

x=0

Explore the idea

Slide left by 1

The graph of (x + 1)|x − 1| is the graph of x|x − 2| moved one unit left, together with its touching and crossing points.

Start: y = x|x − 2|

Touches $y=1$ at $x=1$ and crosses it at $x=1+\sqrt2\approx2.41$ .

Marked points at this step, rounded to 6 significant figures.
Point	x	y
x = 1	1	1
x = 1+√2	2.41421	1

Graph at the selected step
Points where the curve meets y = 1

Ideas to try

Which solution of (a) became

x=0

Split at the absolute-value boundary

Solution

Absolute-value inequalities

Case $x\ge2$

Case $x<2$

Combine

y = x|x − 2| against y = 1

Shifting a known result

Spot the substitution

Use the earlier answer

Solve for $x$