Ch. 4.2 Mean Value Theorem for Derivatives

1. Ch. 4.2 Mean Value Theorem for Derivatives

2. If f (x) is continuous over [a,b] and differentiable over (a,b), then at some point c between a and b: The Mean Value Theorem says that at some point in the closed interval, the actual slope equals the average slope. Mean Value Theorem for Derivatives The Mean Value Theorem only applies over a closed interval.

3. Tangent parallel to chord. Slope of tangent: Slope of chord:

4. Note: Increasing does not mean that the derivative is always positive. Decreasing does not mean that the derivative is always negative. Increasing and decreasing functions: A function is increasing over an interval if: when then . A function is decreasing over an interval if: when then .

5. Even though the derivative is zero, the function is increasing. This is an increasing function because as we move to the right, the y value is always increasing.

6. The terms "increasing" and "decreasing" refer to intervals, not to individual points. Zero can be in both intervals, because both intervals satisfy the definitions of increasing or decreasing. This function is increasing on the interval: This function is decreasing on the interval:

7. Functions with the same derivative differ by a constant. These two functions have the same slope at any value of x.

8. could be or could vary by some constant . Example: Find the function whose derivative is and whose graph passes through . so:

9. Example: Find the function whose derivative is and whose graph passes through . so: Notice that we had to have initial values to determine the value of C.

10. Antiderivative A function is an antiderivative of a function if for all x in the domain of f. The process of finding an antiderivative is antidifferentiation. The process of finding the original function from the derivative is so important that it has a name: You will hear much more about antiderivatives in the future. This section is just an introduction.

11. Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward. (We let down be positive.) Since acceleration is the derivative of velocity, velocity must be the antiderivative of acceleration.

12. Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward. The power rule in reverse: Increase the exponent by one and multiply by the reciprocal of the new exponent. Since velocity is the derivative of position, position must be the antiderivative of velocity.

13. Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward. The initial position is zero at time zero. p

14. Homework: 4.2 4.2 p 202 3,12,21,30 4.1 p 193 13, 23, 37, 46 3.3 p 124 6,12,18,24,30