Math 114, Calculus I, Laboratory 11:
The Mean Value Theorem, Monotonic Functions and Concavity
1.Look at FIGURE 6 page 265 and read EXAMPLE 3 page 264. Open the MapleV procedure titled MVT from your diskette. Enter the command
MVT(x->x^3-x,0,2,-2,6,20);
What this does is to make an animation of a tangent line moving along the graph and shows the slope of the tangent line. In this particular case, the function is the one of EXAMPLE 3, and after some thought the viewing rectangle was set to Ymin=-2 and Ymax=6. The numbers of frames was set to 20. Notice the "VCR control panel" near the top. The most important buttons will be third, fourth and fifth. In the frame displayed below, we see that the slope of the tangent line at x=1.6 is about 6.680, quite far from the mean value of 3.
According to the text, the value of c is . So we should "zoom-in" to investigate the situation here. We set Ymin=f (1.10) and Ymax=f (1.20).
>MVT(x->x^3-x, 1.10 ,1 .20 , 1.10^3-1.10 , 1.20^3-1.20 , 20);
Experiment with the procedure MVT for problems 1-4 on page 266, and use the procedure MVT to estimate all values of c that satisfy the conclusion of the Mean Value Theorem.
SUBMIT Your estimates for the values of c, and printouts of frames supporting your estimates. In each case, turn in a "zoomed-out picture" like the one above, and "zoomed-in pictures"…one for each value of c.
2.Read problem 6 page 266. Try
> MVT(x->(x-1)^(-2) , 0 , 2 , 0 , 300 , 21 );
SUBMIT a paragraph explaining why it might seem like the Mean Value Theorem is contradicted, and explain why there is not contradiction. Use good logic and grammar in your explanation!
- Read EXAMPLE 1 on page 269. In order to find where the function is increasing, we need to solve the equation D(f)(x)=0 for x, so with MapleV we say solve(D(f)(x)=0,x); The equation has three solutions: . We come to the conclusion that the function is increasing on
[-1,0] and on [2,), and that it is decreasing on (-,-1] and [0,5]. Let's produce FIGURE 3 like this:
with(plots):
g1:=plot(f(x) , x=-2..-1,color=red):
g2:=plot(f(x) , x=-1..0,color=green):
g3:=plot(f(x) , x=0..2,color=red):
g4:=plot(f(x) , x=2..3,color=green):
display(g1,g2,g3,g4);
Notice the colons instead of semi-colons.
This emphasizes the parts of the graph where the function is increasing (green) and decreasing (red). Produce analogous graphs for problems #26 and #28 on page 272. Since colors will not show up on a printout, use style=point instead of color=green.
SUBMIT printouts of the two graphs, and work showing how you solved for the critical points.
- Now read EXAMPLE 1 page 275. In order to find out where the function is concave up, we need to set the second derivative equal to zero and solve for x:
solve( D(D(f))(x)=0,x); in order to find the so-called points of inflection. To get FIGURE 5 we use the commands
with(plots):
g1:=plot(f(x) , x=-2..0,color=red):
g2:=plot(f(x) , x=-0..2,color=green):
display(g1,g2);
As in the above activity, this emphasizes the parts of the graph where the function is concave down (red) and concave up (green). Produce analogous graphs for problems #20 and #22 on page 278. Since colors will not show up on a printout, use style=point instead of color=green.
SUBMIT printouts of the two graphs, and work showing how you solved for the inflection points. - Do problem #36 on page 279.
SUBMIT printouts showing how you used MapleV to compute the intervals of concavity to one decimal place, and a graph of f in the spirit of #4 above, showing intervals of downward concavity in style=point, and a couple of sentences explaining why one cannot find the intervals of concavity exactly, and why we need to settle for approximations.