- Topic
- Curve sketching
- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
Curve sketching practice question 23
Let . Find the -coordinate of its local minimum.
is a quotient, so which rule gives ? Once you have the critical points, classify them by the sign of , not by which one is smaller.
Quotient rule: . The critical points are the roots of the numerator.
There are two critical points, one on each side of the vertical asymptote at . Decide which is the minimum from the sign of , or from the shape: the branch to the right of the asymptote comes down from .
An exact expression, or a decimal to two decimal places.
Theory UBC Math 100 §7CLP-1 §3.6Key results
Worked solution
Strategy
Differentiate, solve the quadratic in the numerator, and classify the two critical points. The vertical asymptote separates a local maximum on the left from a local minimum on the right.
Working
Quotient rule.
Quadratic formula.
Falling then rising to the right of the asymptote: a local minimum at 1 + √2 ≈ 2.414 (value 2 + 2√2). Left of the asymptote, f′ > 0 for x < 1 − √2 and f′ < 0 between 1 − √2 and 1, so 1 − √2 is a local maximum.
Answer
The answer is 1 plus root 2, about 2.414.
Check
f(2) and f(3) both equal 5, and f(1 + √2) = 2 + 2√2 ≈ 4.83 dips below them.
Where this goes wrong
Picking 1 − √2 ≈ −0.414 because it is the smaller x, or because f is smaller there. It is a local maximum; the local minimum is the critical point on the right of the asymptote, where the branch turns from falling to rising.
Leans on
The quadratic formula, from Functions, graphs and inverses
Try a similar question
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