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Photo by Vickie Kelly, 2002. Greg Kelly, Hanford High School, Richland, Washington. 4.2. Mean Value Theorem for Derivatives. Teddy Roosevelt National Park, North Dakota. If f ( x ) is continuous over [ a , b ] and differentiable over ( a , b ), then at some point c between a and b :.

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## Mean Value Theorem for Derivatives

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**Photo by Vickie Kelly, 2002**Greg Kelly, Hanford High School, Richland, Washington 4.2 Mean Value Theorem for Derivatives Teddy Roosevelt National Park, North Dakota**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.**Tangent parallel to chord.**Slope of tangent: Slope of chord:**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:**Functions with the same derivative differ by a constant.**These two functions have the same slope at any value of x.**could be**or could vary by some constant . Example 6: Find the function whose derivative is and whose graph passes through . so:**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.**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.**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.**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.**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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