- Topic
- Curve sketching
- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 3
Curve sketching practice question 24
How many real solutions does the equation have?
The solutions are where the curve meets the line . Can you recast that as the zeros of a single function, and let its derivative tell you how many there can be?
Count the zeros of instead. Where is decreasing, where is it increasing, and what is its smallest value?
is zero at . Compare with 0, using , and find the limits of as .
A whole number.
Theory UBC Math 100 §7CLP-1 §3.6Key results
Worked solution
Strategy
Turn the equation into the zeros of one function and sketch it from its derivative: where it falls, where it rises, how low it gets, and what it does at both ends.
Working
Solutions of eˣ = 3x are exactly the zeros of g.
eˣ increases, so it passes 3 only once: g falls, then rises, with its minimum at log 3.
log 3 > 1 because 3 > e.
On the left eˣ → 0 while −3x → +∞; on the right eˣ outgrows 3x.
The graph of g is unbroken: it comes down from +∞ to a negative minimum, crossing the axis once, then climbs back to +∞, crossing once more. On each side of log 3 it moves in one direction only, so it cannot cross twice there. Two solutions.
Answer
The answer is two.
Check
eˣ − 3x is 1 at x = 0, e − 3 ≈ −0.28 at x = 1 and e² − 6 ≈ 1.39 at x = 2, so it changes sign twice.
Where this goes wrong
Deciding from a rough picture that the line y = 3x misses the curve y = eˣ, or counting only one crossing. The minimum of eˣ − 3x is negative, so the line rises above the curve between two crossing points.
Leans on
The derivative of eˣ, from Exponential and trigonometric functions
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