- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 3
Power functions and first sketches practice question 9
The speed of a reaction depends on the concentration of a chemical according to the function below. As grows large, approaches a limiting value. Find the concentration at which is exactly half of that limiting value.
Two steps: first the limiting value, then the that gives half of it. For huge , which terms of the fraction actually matter?
For large , the in the denominator is negligible next to . What does the fraction become?
Half of the limiting value is . Solve .
An exact expression, or a decimal to three decimal places.
Theory UBC Math 100 §1.2Key results
Worked solution
Strategy
First find the limiting value by keeping only the dominant terms for large x. Then set y equal to half of it and solve for x.
Working
In the denominator, x³ dominates the constant 2. The limiting value is 6.
Half of 6 is 3; multiply both sides by the denominator.
Collect the x³ terms and take the cube root. x³ is increasing, so x³ = 2 has exactly one real solution, ∛2, and it is positive, as a concentration must be.
Answer
The answer is the cube root of 2, about 1.260.
Check
At x = ∛2, x³ = 2 and y = 12/4 = 3, exactly half of the value 6 that y approaches.
Where this goes wrong
Taking the limiting value to be 3 because “6 divided by 2”. The 2 is added to x³, not multiplied; for large x it is swamped, and the fraction tends to 6.
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