Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
4. Applications of Derivatives
Differentials
Problem 53b
Textbook Question
Mean Value Theorem The population of a culture of cells grows according to the function P(t) = 100t / t+1, where t ≥ 0 is measured in weeks.
b. At what point of the interval [0, 8] is the instantaneous rate of change equal to the average rate of change?
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1
First, calculate the average rate of change of the population function P(t) over the interval [0, 8]. This is done using the formula: (P(8) - P(0)) / (8 - 0).
Next, evaluate P(0) and P(8) by substituting these values into the function P(t) = 100t / (t + 1).
After finding P(0) and P(8), compute the average rate of change using the values obtained.
Now, find the derivative of the function P(t) to determine the instantaneous rate of change. Use the quotient rule for differentiation since P(t) is a quotient of two functions.
Finally, set the derivative equal to the average rate of change calculated earlier and solve for t in the interval [0, 8] to find the point where the instantaneous rate of change equals the average rate of change.
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