O.K. But what if the chicken didn t have access to a teleporter.

Transcription

1 The intermediate value theorem, and performing algebra on its. This is a dual topic lecture. : The Intermediate value theorem First we should remember what it means to be a continuous function: A function f(x) is continuous at some number a if: A function is just called continuous if Today we introduce an important property of continuous functions: The Intermediate Value Theorem. Let us first illustrate it via an example. Suppose that as you drive along you notice a chicken on the east side of a road. On the way back you see the same chicken on the west side of the road. Does it follow that the chicken was at some moment in the exact middle of the road? O.K. But what if the chicken didn t have access to a teleporter. Lets see the same topic now said slightly more abstractly. If you have a continuous function f(x) and a number M and you have f(a) < M for some a and f(b) > M for some b then you have a graph like (a, f(a) (b, f(b) y = M Is it possible to get from the left part of the graph to the right continuously without crossing the line y = M? Does f(x) = M have a solution? This is exactly what the intermediate value theorem says:

2 2 Theorem (The intermediate value theorem). Let f(x) be a continuous function. Let a, b, and M be numbers. Suppose that f(a) < M and f(b) > M. Then there is a number c between a and b for which f(c) = M Using the IVT x 3 + 3x is continuous. Let s say that you want to know whether p(x) = has a solution. First Try p(0) =. How does this compare to? Second Try p() =. How does this compare to? So we have that p(0) p(). Is p(x) continuous? And by the intermediate value theorem The general process for these problems: If you want to show that f(x) = M has a solution: () Check if M is continuous. If it has some discontinuities, then make sure that you avoid them. (2) Find two inputs a and b for which f(a) < M < f(b). Just try numbers until it works out. (3) Is f(x) continuous on [a, b]? If so, you can now use the IVT to conclude that f(x) = M has a solution. Let p(x) = x 4 + x 3 + x 2 + x + Show that p(x) = π has a solution. Show that p(x) = π has a solution. Using algebra when computing its. Lots of times continuity makes its easy to compute. All you need to do is plug in the value where you want to take the it. x 3 x 2 Sometimes this idea results in something which makes no sense and we have to do some algebra before we can even take the it: x 2 9 Think about. What do you get when you plug in x = 3? Does it make any x 3 x 3 sense?

3 If you factor the numerator and denominator, then something cancels. Try to take the it again: x 2 9 x 3 x 3 3 ( The strategy: If you compute ) a it and get something which doesn t make any sense 0 0,,, 0,... then try to do some These are called indeterminate forms. When you get its of this form you need to try to do some algebra and simplify. What to do when you see 0 0 A strategy: Try to find a factor in the numerator and denominator which makes each zero. Cancel them: x 2 x 2 x 2 x 2 3x + 2 x π tan(x) sin(x) Before we introduce how s in the it can make things hard, lets talk about when they don t make things hard. Performing algebra with. All of the it laws from last week still work with infinite its, with the added rules: () If c is a number then c + = and c = (2) If c 0 is a number then c = (3) If c > 0 is a number then c = and c = (4) If c < 0 is a number then c = and c =

4 4 (5) positive 0 x c f(x) = number?) (6) negative 0 x c f(x) = If f(x) = 0 and f(x) > 0 for x close to c then x c (What happens when you divide by a tiny positive If f(x) = 0 and f(x) < 0 for x close to c then x c x + (x ) x 0 + x Indeterminate forms: There is no hard and fast trick which will always make indeterminate forms better. A good strategy is to just try some algebra until it stops being indeterminate. What to do when you see A strategy: Find the term in the numerator (or denominator) which makes it infinite. Divide the numerator and denominator by this. x 2 x 2 x 0 x 3x x 0 x 2 x 2 x 3x x π tan(x) x sin(x) +

5 5 : Try to combine the terms together 4 x 2 + x 2 x 2 4 Another strategy for indeterminate forms Multiplying by the conjugate. x x x Multiply and divide by + x (this is called the algebraic conjugate) x x = Does this look any nicer? So we need to compute x x x = x 2 x 4 x 4 h 0 (2 + h) h