AP Calculus Unit 4 Day 5 Finish Concavity Mean Value Theorem Curve Sketching

Practice Given f(x) = x 3 – 6x 2 – 36x + 3 1)Create a f’(x) sign chart. 2)Where is the function increasing, decreasing, local max and local mins? 3)If the function above is bounded by [0,5] find absolute max and absolute min.

is positive Curve is concave up. is negativeCurve is concave down. is zero Possible inflection point (where concavity changes). What can the 2 nd Derivative tell us? Location of minimum Location of maximum

Applying the Definition of Concavity Free Response Type Question

1.Graph to answer the question: 2.Justify the answer using Calculus (Definition of Concavity): a.Must show is decreasing, which means must show 3.Interpret your work: Since is negative on, is decreasing which means is concave down on this interval based on the definition of concavity. Applying the Definition of Concavity Free Response Type Question

SAMPLE Free Response Type Question:

1.Graph to answer the question: 2.Justify the answer using Calculus (Definition of Concavity): a.Must show is increasing on the interval, which means must show on the interval 3.Interpret your work: Since is positive on, is increasing which means is concave up on this interval based on the definition of concavity.

Definition – Point of Inflection

This graph illustrates concavity change and no tangent line at that point of change. Not a Point of Inflection, Even though concavity changed. What is the significance of the “tangent line” requirement?

An example of : What is the significance of the “tangent line” requirement?

To locate POSSIBLE inflection points Determine what x values for which f’’(x) = 0 or f’’(x) does not exist

Example (continuing with same function): Finding the inflection points: Set the second derivative equal to zero. Possible inflection point at. negative positive inflection point at

What do the different functions tell you? PUT this CONCEPT table in your notes!!! ff 'f ’’ PositionVelocity (change in position) Acceleration (change in velocity) Zeros, asymptotes, gaps Where f is increasing or decreasing Where f ‘ is increasing or decreasing Locations of possible extrema Locations of possible points of inflection “critical points”Concavity

Next on the AGENDA.... Curve Sketching A useful skill for some students Sometimes, but not often, could appear as a free response question Regardless, conceptual knowledge that is helpful

f(2)=2, f(0)=-4 and f(-3)=4 KNOWN POINTS! PLOT THEM f ’(x)=0 at x=0 and x=-3 CRITICAL POINTS (POSSIBLE EXTREMA) LABEL THEM f ’(x)>0 for x>0 and x<-3; f ’(x)<0 for -3<x<0 INTERVALS OF INCREASING/DECREASING FOR f(x). ***Might help to make a sign line for f’(x)*** f ‘’(x)>0 for x>-1.5 and f ‘’(x)<0 for x<-1.5 INTERVALS OF CONCAVE UP/DOWN FOR f(x). ****Make a sign line for f ‘’(x).**** Sketch the graph of a continuous function f(x) that meets the following criteria. SUGGESTED “TOOL” Sign Charts for 1 st and 2 nd Derivatives.

Additional practice that will be posted on Blackboard for you to reference TAKE note (i.e. WRITE down) the solution Next two slides...

Today’s Theorem.. Mean Value Theorem (Reference the Definitions & Theorems powerpoint printout.)

Mean Value Theorem (MVT) If f is continuous on [a,b] and differentiable on (a,b), then there exists a number, c, in (a,b) such that

If the MVT is applicable find the value of “c” such that Practice

Practice--ANSWERS

Find the point(s) guaranteed to exist by the mean value theorem on [-1,2] for the equation Remember: The theorem requires the function must be continuous on closed interval [a,b] but values guaranteed to exist are in open interval (a,b)

Rolle’s Theorem (A special case of MVT) If f is continuous on [a,b] and differentiable on (a,b) AND f(b)=f(a) Then there is at least on c in (a,b) such that f’(c)=0. ***Therefore there is guaranteed to exist a horizontal tangent at some point on f(x).***

Use the mean value theorem to explain why there is guaranteed to exist a horizontal tangent at some location on [-3,1] for the following curve.

Function f is continuous on [0,10] and differentiable on (0,10). Given the following data, what is the minimum number of extrema that must exist on the interval (0,10)? Justify. X f Answer: There are at least two extrema on (0,10). One in the interval (0,4) and one in the interval (4,10). Extrema occur at critical points, where the 1 st derivative equals zero. The average slope on (0,4) is zero. There is guaranteed to exist a point in (0,4) where f’(0) = 0 because of the Mean Value Theorem. The same reasoning guarantees the existence of an extrema in the interval (4,10).

Function f is continuous on [6,10] and differentiable on (6,10). Given the following data, what is the minimum number of extrema that must exist on the interval (6,10)? Justify. Answer: There is at least one extrema on (6,10). Extrema occur at critical points, where the 1 st derivative equals zero. The Intermediate Value Theorem guarantees that f(x) = 3 somewhere in the intervals (6,8) and (8,10). The average slope between those two points is zero. The Mean Value Theorem guarantees another point in that interval where f’(x) = 0. X6810 f425 Note: Picking 3 as the value for f(x) was arbitrary. The key is to find a function value that you can guarantee exists in the two intervals.