Desperate help with mvt !

[abedkyle]

Help with this would be greatly appreciated ! Thanks ! I HAVE NO IDEA HOW TO DO IT!

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 "qualitative" and "quantitative" information)!

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 and the information obtained in previous parts of this problem.

 

[stapel]

I HAVE NO IDEA HOW TO DO IT!

A good place to start would be with your knowledge of graphing and slopes (from back in algebra) and your knowledge of derivatives (from calculus). Also, use the hints they suggest. To start:

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.

What is the slope (approximately) between x = -2 and x = -1? Between x = -1 and x = 0? What then MUST be happening to the value of f(x)? Given the initial value at x = -2, where MUST the graph be going?

 

[HallsofIvy]

The problem says "using the MVT". The Mean value theorem says that, as long as f is continuous on [a, b] and differentiable on (a, b), there exist some c in [a, b] such that f(b)- f(a)= f'(c)(b- a). You are asked to prove that -2< f(0)< -1. The one piece of information you have about values of f is that f(-2)= 1.

So apply the mean value theorem to (-2, 0). f(-2)- f(0)= f(-2)- 1= f'(c)(-2- 0)= -2f'(c). What is the largest value of f' between -2 and 0? What is the smallest?