2. What is the mean
value theorem?
The mean value theorem, often abbreviated as
MVT, is a fundamental concept in calculus. It
serves as a bridge between the average rate of
change of a function and its instantaneous rate of
change. In essence, the MVT tells us that if a
function is continuous on a closed interval [a, b]
and differentiable on the open interval (a, b), then
at some point 'c' within the open interval, the
instantaneous rate of change (derivative) of the
function is equal to the average rate of change
over the closed interval [a, b]
3. Applying the mean value theorem
Example problem 1
Scenario: Imagine a car is traveling
along a straight road, and you want
to calculate its average and
instantaneous velocity during a
specific time interval
Velocity of a moving car
Given:
■ Initial time, t1 = 2 seconds
■ Final time, t2 = 6 seconds
■ Initial position, s(t1) = 10 meters
■ Final position, s(t2) = 50 meters
4. Step-by-step solution
What is the average and instantaneous velocity of the car between t1 and t2?
■ Average velocity = (change in position) / (change in time)
■ Average velocity = [s(t2) - s(t1)] / [t2 - t1]
■ Average velocity = [50 m - 10 m] / [6 s - 2 s] = 40 m/s / 4 s = 10 m/s
Step 1: calculate average velocity
1
5. Step-by-step solution
What is the average and instantaneous velocity of the car between t1 and t2?
■ Let v(t) represent the velocity of the car
■ According to the MVT, there exists a time 'c' between t1 and t2 where v'(c) =
[v(t2) - v(t1)] / [t2 - t1]
■ We have the average velocity (10 m/s) as calculated above
■ Therefore, v'(c) = 10 m/s
The average velocity of the car between 2 seconds and 6 seconds is 10 meters per
second. By the mean value theorem, there exists a time 'c' within this interval
where the instantaneous velocity is also 10 meters per second
Step 2: apply the MVT to find instantaneous velocity
2
6. Applying the mean value theorem
Example problem 2
Scenario: You have a function f(x) = x^3 - 3x^2 + 2 on the interval [0, 3]. Prove the
existence of at least one point 'c' within this interval where the instantaneous rate of
change (derivative) equals the average rate of change
Proving the existence of a turning point
7. Step-by-step solution
Example problem 2
■ The function f(x) is continuous on the closed interval [0, 3]
■ The function f(x) is differentiable on the open interval (0, 3)
Step 1: check MVT conditions
1
8. Step-by-step solution
Example problem 2
■ Average rate of change = [f(3) - f(0)] / (3 - 0)
■ Average rate of change = [(3^3 - 3(3^2) + 2) - (0^3 - 3(0^2) + 2)] / 3
■ Average rate of change = [27 - 27 + 2] / 3 = ⅔
Step 2: calculate average rate of change
2
9. Step-by-step solution
Example problem 2
■ According to the MVT, there exists at least one point 'c' in the interval
(0, 3) where f'(c) = average rate of change
■ We need to find 'c' such that f'(c) = 2/3
Step 3: apply the MVT
3
10. Final conclusion
Example problem 2
By the mean value theorem, there exists at least one point 'c' within the interval (0,
3) where the instantaneous rate of change (derivative) of f(x) is equal to the
average rate of change, which is 2/3
Final conclusion
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