Applying the Mean Value Theorem

soe236
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Hey,
I know the basic definition of the MVT, but I'm having a lot of trouble applying it to this problem. I would greatly appreciate any kind of help or guidance.

A graph of the derivative of f(x) is displayed below. Information about the function f(x) is known only for -2.5 < x < 3.5. Also f(-2) = 1. Consider the graph carefully, and consider the information in both the numbers and the shapes of the graph (both "quantitative" and "qualitative" information).

the graph:
http://img148.imageshack.us/img148/5224/graphfx6.jpg

a) Explain why -2 < f(0) < -1. Look carefully at the graph and make estimates using the MVT. Explain the steps of your reasoning in detail.
b) Explain why f(3) > 4 +f(1). Again, use the MVT and explain your reasoning in detail.
c) How big and how small can f(1) - f(0) be?
d) Use the information in a), b), and c) to explain why f(3) must be positive.
e) Explain why f(x) = 0 must have a solution between 0 and 3. Use the IVT (intermediate value th.) and the
information obtained in previous parts of this problem.
 
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Starting from x = -2, how does f(x) behave, can you tell by looking at f'(x)?
 
EnumaElish said:
Starting from x = -2, how does f(x) behave, can you tell by looking at f'(x)?

well from x=-2 to about x=.5, f'(x) is negative which means f(x) is decreasing and f(x) is increasing on approx [0.5,3.5] ..do i do something with the first derivative test? I just don't know what the question is asking for or atleast what I have to find to answer it
 
and the critical point is at 0.5 if that's of any use?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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