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Caeder
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Let [tex]f(x) = \frac{x}{x-1}[/tex]. Prove [tex]f(x)[/tex] is uniformly cont. on the interval [tex][1.5,\infty)[/tex]
Caeder said:Let [tex]f(x) = \frac{x}{x-1}[/tex]. Prove [tex]f(x)[/tex] is uniformly cont. on the interval [tex][1.5,\infty)[/tex]
rs1n said:Hint: What can you say about f'(x). You can prove this in almost one line of math.
Caeder said:Uh, [tex]\frac{-1}{(x-2)^2}[/tex] ... I don't see how f'(x) helps in showing f(x) is uniformly continuous. We know every function that is integrable is continuous, but we can't say the same with differentiability. Mind helping?
rs1n said:I think perhaps the hint may be too subtle. Honestly, I'm trying to avoid giving the answer as this seems like an analysis h.w. problem. Think 'bounded,' and look up 'Lipschitz.' At least this way you're doing some of your own "research" on the problem.
P.S. Unless I'm mistaken, your derivative is incorrect if we are to assume your problem statement is correct. Check your denominator.
[tex]f(x) = \frac{x}{x-1} = \frac{x-1}{x-1} + \frac{1}{x-1} = 1 + (x-1)^{-1} \implies
f'(x) = -\frac{1}{(x-1)^2}[/tex]
Caeder said:Indeed, I meant 1. Typo. I'll look for "Lipschitz" although it doesn't sound familiar.
rs1n said:If Lipschitz doesn't ring a bell, the mean value theorem should.
Caeder said:Well, looking in my notes, to show [tex]f(x) = \frac{1}{x}[/tex] is unif. continuous on [tex]A = [a, \infty)[/tex] for [tex]a > 0[/tex], they prove it with:
We note [tex]x,y \in A[/tex], then:
[tex]|f(x) - f(y)| = \left|\frac{y-x}{xy}\right| \leq \frac{|y-x|}{a^2}[/tex]
Here, if we have that [tex]\epsilon > 0[/tex] and if [tex]|x-y| < a^2\epsilon[/tex], then:
[tex]|f(x) - f(y)| < \epsilon[/tex]
Hence f is unif. continuous.
...I see no use of derivative. I tried manipulating mine to be similar to the above one, but I'm not even enitrely sure of how the above proof works. Sorry to be a pain.
Caeder said:Uh, [tex]\frac{-1}{(x-2)^2}[/tex] ... I don't see how f'(x) helps in showing f(x) is uniformly continuous. We know every function that is integrable is continuous, but we can't say the same with differentiability. Mind helping?
rs1n said:First, let's see what they do:
[tex]f(x)-f(x) = \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy} = (y-x) \cdot \frac{1}{xy}[/tex]
Notice that
[tex]f(x) = \frac{1}{x} \le \frac{1}{a}[/tex]
The rest follows immediately.
There is an easier way to do that same problem using the mean value theorem. The method above works, but it doesn't generalize easily. Here's the general method (and I'll only sketch the proof).
Your function is clearly continuous on [tex][1.5,\infty)[/tex]. Apply the mean value theorem:
[tex]\frac{f(y)-f(x)}{y-x} = f'(c) \text{ where } c \in (x,y) [/tex]
Toss in absolute values on both sides.
[tex] \left| \frac{f(y)-f(x)}{y-x}\right| = |f'(c)|
\implies |f(y) - f(x)| = |f'(c)| \cdot |y-x|[/tex]
Again, what do you property do notice about the derivative on the interval [tex][1.5,\infty)[/tex]? To answer this question, think about what you would need to make the rest of this sketch work? Once you figure this out, you should be able where the derivative "appears" in your example problem (it has to do with the [tex]1/a^2[/tex]).
That's exactly backwards! If a function is differentiable, then it is continuous. An integrable function may not be continuous.Caeder said:Uh, [tex]\frac{-1}{(x-2)^2}[/tex] ... I don't see how f'(x) helps in showing f(x) is uniformly continuous. We know every function that is integrable is continuous, but we can't say the same with differentiability. Mind helping?
HallsofIvy said:That's exactly backwards! If a function is differentiable, then it is continuous. An integrable function may not be continuous.
The point in the orginal response was that if a function is continuous on a closed interval, then it is uniformly continuous there. [1.5, [itex]\infty[/itex]) is closed and x/(x-1) is continuous on it.
Caeder said:We know that the derivative is always negative for values > 1.
So we have:[tex]f(y) - f(x) \rightarrow \frac{\frac{y}{y-1} - \frac{x}{x-1}}{y-x} = \frac{-1}{(x-1)(y-1)}[/tex]... Not sure what this quite does... we know this equals |f'(c)|*|y-x| from what you showed me. f'(c) will have to be 1.5 or greater, and thus this is negative. Hence this product will be positive assuming y > x...
Uniform continuity is a mathematical concept that describes the behavior of a function over its entire domain. A function is uniformly continuous if, for any given value of epsilon, there exists a corresponding value of delta such that the distance between the outputs of any two points within delta distance of each other is less than epsilon distance apart.
Uniform continuity is a stronger condition than regular continuity. While regular continuity only requires that the function be continuous at every point within its domain, uniform continuity requires that the function be continuous over its entire domain. This means that the function's behavior must be consistent and predictable across its entire domain, not just at individual points.
If a function is uniformly continuous, it means that small changes in the input will result in small changes in the output. This makes the function more stable and predictable, which can be useful in various applications, such as in engineering and physics.
In order to prove that a function is uniformly continuous, you must show that for any given epsilon, there exists a corresponding delta such that the distance between the outputs of any two points within delta distance of each other is less than epsilon distance apart. This can be done using the formal definition of uniform continuity, along with algebraic manipulations and logical reasoning.
No, not all continuous functions are uniformly continuous. While all uniformly continuous functions are also continuous, the reverse is not necessarily true. This is because uniform continuity requires a stronger and more specific type of continuity, which not all continuous functions possess.