400 likes | 408 Views
Understand Rolle's Theorem and Mean Value Theorem, explore inflection points, concavity, and antiderivatives, and learn about functions from derivatives in calculus. Study examples and important concepts in derivatives and antiderivatives.
E N D
5.2…and...5.3 All mixed together….
Rolles Theorem If f is continuous on [a,b] and differentiable on (a,b) and if f(a)=f(b), then f ‘(c)=0 for at least one number c in (a,b).
Mean Value Theorem (MVT) If y = f(x) is continuous on [a,b] and differentiable on (a,b), then there is at least one point c in (a,b) such that
Tangent parallel to chord. Slope of tangent: Slope of chord:
Inflection Point A point where the graph of a function has a tangent line and where the concavity changes is a point of inflection.
is positive is negative is zero is positive is negative is zero First derivative: Curve is rising. Curve is falling. Possible local maximum or minimum. Second derivative: Curve is concave up. Curve is concave down. Possible inflection point (where concavity changes).
Example: Graph
rising, concave down local max falling, inflection point local min rising, concave up Make a summary table: p
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.
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.
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