- Topic
- Optimization
- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Optimization practice question 3
Find the minimum value of for .
There are no endpoints to check on . Where, then, can the smallest value of occur?
. Set it to zero and keep only the positive solution.
A whole number.
Theory UBC Math 100 §8CLP-1 §3.5Key results
Worked solution
Strategy
On an open interval there are no endpoints to check, so the minimum must be at a critical point; confirm it is a minimum with the second derivative, then evaluate.
Working
The only critical point in the domain.
Concave up throughout, so the critical point is the global minimum on (0, ∞).
Minimum value.
Answer
The answer is 6.
Check
x + 9/x − 6 = (x − 3)²/x, which is never negative for x > 0, so g never drops below 6.
Where this goes wrong
Answering 3, the location of the minimum, instead of the value g(3) = 6. As x → 0⁺ and x → ∞ the function grows without bound, so 6 really is the least value.
Leans on
The power rule with negative exponents, from Differentiation rules
Try a similar question
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