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Mean Value Theorem for Derivatives

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1 Mean Value Theorem for Derivatives
4.2 Mean Value Theorem for Derivatives Teddy Roosevelt National Park, North Dakota Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2002

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

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

4 A couple of somewhat obvious definitions:
A function is increasing over an interval if the derivative is always positive. A function is decreasing over an interval if the derivative is always negative.

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

6 Example 6: Find the function whose derivative is and whose graph passes through so: could be or could vary by some constant .

7 Example 6: 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.

8 The process of finding the original function from the derivative is so important that it has a name:
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. You will hear much more about antiderivatives in the future. This section is just an introduction.

9 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.

10 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.

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. The initial position is zero at time zero. p


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