1. If f (x) is a differentiable function over [ a, b ], then at some point between a and b : Mean Value Theorem for Derivatives

2. If f (x) is a differentiable function over [ a, b ], then at some point between a and b : Mean Value Theorem for Derivatives Differentiable implies that the function is also continuous.

3. If f (x) is a differentiable function over [ a, b ], then at some point between a and b : Mean Value Theorem for Derivatives Differentiable implies that the function is also continuous. The Mean Value Theorem only applies over a closed interval.

4. If f (x) is a differentiable function over [ a, b ], then at some point 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.

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

6. 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:

7. f(x) = x 3 - 4x 2 + 5 f‘(x) = 3x 2 – 8x What relationship is there between this graph of a function and its derivative?

8. Pg. 192, #1 Find any maximums or minimums. And the intervals on which the function is increasing and decreasing. f(x) = 5x – x 2

10. #15 Find each value of c in (a, b) that satisfies the Mean Value Theorem for Derivatives. f(x) = x 2 + 2x – 1, D:[0, 1]

11. #15, f(x) = x 2 + 2x – 1, D:[0,1]

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

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

14. 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.

15. 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.

16. Antiderivative The power rule in reverse –1. Increase the exponent by 1 –2. Multiply by the reciprocal of the new exponent f(x) = f‘(x) = 3x 2 - 4

17. Example Find the equation of a function whose derivative is f’(x) = x 2 + 2x + 1 and passes through the point (1, 3).

18. Example (2) Find the equation of the function whose derivative is f’(x) = 3x 3 – 2x + 1 and passes through the point (2, -1).

19. 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.)

20. 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.

21. 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.