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madgab89
Nov25-08, 06:54 PM
1. The problem statement, all variables and given/known data
Figure 4.46 shows the second derivative of h(x) for -2 \leq x \leq 1

a) Explain why h'(x) is never negative on this interval.
b) Explain why h(x) has a global maximum at x=1
c) Sketch a possible graph of h(x) for this interval.


I realize this is probably a fairly simple question, however it's just making my head hurt.

http://i414.photobucket.com/albums/pp222/madgab89/Picture2.jpg

D H
Nov25-08, 07:42 PM
Are you sure you have given us all of the data? Perhaps the value of h'(x) at x=-1?

madgab89
Nov25-08, 08:12 PM
sorry, don't know how i could have missed that

h'(-1)=0
h(-1)=2

D H
Nov25-08, 09:01 PM
Much better.

The relation between the first and second derivative is that the second derivative of some function is the first derivative of the first derivative of that function.

So, to avoid this confusion, for a while let's denote h'(x) as g(x). Then g'(x)=h''(x). Now, if I showed you that graph in the first plot and labeled it as g'(x) rather than h''(x) and asked you to explain why g(x) is never negative on the interval in question, could you do that?

HallsofIvy
Nov26-08, 04:50 AM
For x\ge -1, the graph shows that h"(x) is positive. That means h'(x) is increasing. And since h'(-1)= 0 ....


Now that you know that h' is positive on the interval, h(x) is increasing so its minimum and maximum on [-1, 1] must be at ...