Sample Problems, 4.3

Sample Problems, Lesson 4.3

Example Problem. #2, Lesson 4.3

	Given Problem. #2, Lesson 4.3
Part (a). Intervals in which the pictured function is concave up: (-1,2), (7,8)	Here we simply observe where the curve is concave up or opening up.
The largest of these is (-1,2)	This interval is 3 units "wide" while the other is only 1 unit "wide." This is the answer to part (a).
Part (b). Intervals in which the pictured function is concave down: (2,4), (4,7)	Here we simply observe where the curve is concave down or opening down.
The largest of these is (4.7)	This interval is 3 units "wide" while the other is only 2units "wide." This is the answer to part (b).
Part (c). The only point of inflection appears to be: (2,2)	This again is done simply by observation. We are looking for any points where the curve changes from concave down to concave up. (or the reverse) This is the answer to part (c).

Example Problem: #8, Lesson 4.3

Given that ,
(a) Find the intervals on which f is increasing or decreasing.
(b) Find the local maximum and minimum values of f.
(c) Find the intervals of cancavity and the inflection points.

Given Problem. #8, Lesson 4.3

When a function is increasing, its first derivative (slope of tangent line) must be positive, and when a function is decreasing, its first derivative must be negative. So we calculate the 1st derivative.

To Find when the function is increasing, we set the 1st derivative to be greater than zero and solve. We now see that the function is increasing whenever x < 1.

In a similar way, we find when the funciton is decreasing by finding out when the 1st derivative is less than zero. We can see from the work at the left that the funciton is decreeasing whenever x > 1. Part (a) is now complete.

By setting the derivative equal to zero, we can find where there might be an relative maxima or minima. We find that there is one such point, at x = 1. Because f ''(1) is positive, we know that this must be a relative maximum. Part (b) is now complete.

To examine points of inflection, we need to find when the second derivative is equal to zero, so we calculate f ''(x).

We see that there is one point of inflection at x = 0.

Because we know that there is one inflection point at x = 0, we need to examine concavity on either side of this point.Sample points can be chosen to do this. I chose x = -1, and x = 1. Because both tests come out negative, we find that the intervals of concavity are as follows:

Interval	Concavity
	concave down
	concave down

Part (c) is now completed.

Now is a good time to look at a graph of the function for checking purposes. If you have Maple available to you, the command to plot this function would be:

> plot(-x^8+8*x+1,x=-2..2);

Example Problem: #22, Lesson 4.3

Given that , (a) Find the intervals of increase or decrease, (b) Find the local maximum and minimum values, (c) Find the intervals of concavity and the inflection points, Check your work with some form of technology.	Given Problem: #22, Lesson 4.3
	When a function is increasing, its first derivative (slope of tangent line) must be positive, and when a function is decreasing, its first derivative must be negative. So we calculate the 1st derivative.
	To Find when the function is increasing, we set the 1st derivative to be greater than zero and solve. We now see the two times that the function is increasing. The interval of increase is: .
	In a similar way, we find when the funciton is decreasing by finding out when the 1st derivative is less than zero. We can see that there are two intervals where the function is decreasing. That intevals are: and
	By setting the derivative equal to zero, we can find where there might be an relative maxima or minima. We find that there are two points of this type.
	To examine points of inflection, we need to find when the second derivative is equal to zero, so we calculate f ''(x). We can also see from the second derivative that 2nd derivative at is equal to a positive 4, Therefore the function has a local minimum at . (positive 2nd derivative indicates concave up). The second derivative at is negative 4. Therefore the function has a local maximum at . (negative 2nd derivative indicates concave down).
	We see that there is one point of inflection at x = . To the left of this value we can see that the curve is concave up, and to the right of this value the curve is concave down.
	We see the local maximum and local minimum appear to be correct. The point of inflection also appears accurate. We can also check concavity intervals and intervals of increase and decrease. The above analysis does not show the asymptotes or the x-intercept, but calculus greatly helped us to find out about the behaviour of this function.

	*Assignment*
1-27, Odds
Lesson 4.3, Pages 288-290