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Mean Value Theorem for Derivatives4.2 Teddy Roosevelt National Park, North Dakota Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2002

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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 only applies over a closed interval. The Mean Value Theorem says that at some point in the closed interval, the actual slope equals the average slope.

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

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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. A couple of somewhat obvious definitions:

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

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

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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.

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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.

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Since acceleration is the derivative of velocity, velocity must be the antiderivative of acceleration. Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec 2 and an initial velocity of 1 m/sec downward. (We let down be positive.)

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Since velocity is the derivative of position, position must be the antiderivative of velocity. Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec 2 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.

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Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec 2 and an initial velocity of 1 m/sec downward. The initial position is zero at time zero.

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5.3 Definite Integrals and Antiderivatives. 0 0.

5.3 Definite Integrals and Antiderivatives. 0 0.

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