- Topic
- Optimization
- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Optimization practice question 1
Find the global minimum value of on the closed interval .
On the closed interval , where can the smallest value of occur? List every candidate before you evaluate anything.
is zero at , and only lies in . Compare there with and .
An integer.
Theory UBC Math 100 §8CLP-1 §3.5Key results
Worked solution
Strategy
The closed-interval method: list the critical points inside the interval and the two endpoints, evaluate f at each, and pick the smallest.
Working
Critical points.
Candidates.
Smallest of the three.
Answer
The answer is negative 1, attained at x = 1.
Check
f(x) + 1 = (x − 1)²(x + 2), which is never negative for x ≥ 0, so f never drops below −1 on [0, 2].
Where this goes wrong
Answering x = 1, the location, when the question asks for the minimum value f(1) = −1. Read whether a question wants where or what.
Leans on
Finding critical points, from Curve sketching
Try a similar question
Find the absolute maximum value of on the closed interval .Saved in this browser only.