Cópia de EN Mean Value Theorem by Slidesgo.pptx

augustomelo25 16 views 31 slides Sep 26, 2024
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About This Presentation

Mean value theorem


Slide Content

Mean Value Theorem Here is where the lesson begins

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]

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

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

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

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 P roving the existence of a turning point

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

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

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

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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