- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 3
Functions, graphs and inverses practice question 13
How many real solutions does the equation below have?
Solutions are the places where two graphs cross. Which two graphs, and how far from the origin can a crossing possibly be?
Sketch both sides on one set of axes: has amplitude 3 and period , and is a line through the origin.
Since , every solution has . Both sides are odd functions, so count the solutions with , double that, and add .
A whole number.
Theory UBC Math 100 §1.5CLP-1 §0.4Key results
Worked solution
Strategy
Compare a bounded, periodic function with a line. The amplitude confines every solution to |x| ≤ 3, a sketch of the arches in that window shows the crossings, and oddness supplies the negative side.
Working
The line leaves the band −3 ≤ y ≤ 3 beyond x = ±3, and the sine curve never does.
Both sides are 0 at the origin.
For small x, sin(2x) ≈ 2x, so 3 sin(2x) ≈ 6x starts above the line; at x = π/2 the arch is back to 0, below the line, so they cross at least once. On this arch 3 sin(2x) is concave down and the line is straight, so after leaving the origin above the line it can cross only once.
On [π/2, π] the sine curve is not positive, and π > 3, so no further arch fits in the window.
Both sides are odd, so the crossing near x ≈ 1.34 has a mirror image near −1.34. With x = 0 that makes three.
Answer
The answer is three solutions: x = 0 and one on each side of it.
Check
3 sin(2x) − x is about 1.73 at x = 1 and −1.08 at x = 1.5, so a positive root lies between, mirrored by a negative one, and 0 is the third.
Where this goes wrong
Counting only x = 0, or only the solutions with x ≥ 0. The steep start of 3 sin(2x) puts it above the line just right of 0, which forces a crossing in the first arch, and oddness mirrors it to the left.
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