Given #f(x) = x^33x#, how can you construct an infinitely differentiable oneone function #g(x):RR>RR# with #g(x) = f(x)# in #(oo, 2] uu [2, oo)#?
This makes
(All integrals by Wolfram Alpha) And
Never mind.
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Here is my first attempt. (Unfortunately, it fails.)
graph{x^33x [5.454, 5.645, 2.606, 2.94]}
Determined to find a polynomial, I set
But for infinite differentiability, we also need
This attempt fails to satisfy those.
Perhaps we can start with and odd, fourtimes integrable function and work backwards from there.
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See below.
For practical purposes and with a guarantee of accuracy, we can proceed, according to
This for
Solving this system we obtain
So the binding function is
Attached a plot showing the binding.
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A classic example of a function for which the Taylor series fails is:
Inspired by this, consider the function:
graph{e^(x^2/(x^24)^2) [5, 5, 2.5, 2.5]}
Define
graph{x^33x+e^(x^2/(x^24)^2)*(x^9/256+3xx^3)(1+x^2/(x^24)^2+x^4/(2(x^24)^4)) [5, 5, 2.5, 2.5]}
The additional trick here is that the multiplier:
It remains to show that the resulting function is monotonically increasing.
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To construct an infinitely differentiable onetoone function ( g(x): \mathbb{R} \rightarrow \mathbb{R} ) with ( g(x) = f(x) ) in ( (\infty, 2] \cup [2, \infty) ), you can use the following steps:

Define a function ( h(x) ) as follows: [ h(x) = \begin{cases} f(x) & \text{if } x \leq 2 \ f(x) & \text{if } x \geq 2 \ \end{cases} ]

Next, ensure that ( h(x) ) is differentiable in its domain ( (\infty, 2] \cup [2, \infty) ). The function ( h(x) ) is continuous everywhere since it's made up of continuous pieces.

Now, let's construct a smooth connection between the pieces at ( x = 2 ) and ( x = 2 ). Define a function ( k(x) ) such that: [ k(x) = \begin{cases} 0 & \text{if } x \leq 2 \ \text{smooth transition} & \text{if } 2 < x < 2 \ 0 & \text{if } x \geq 2 \ \end{cases} ]

The function ( k(x) ) should smoothly transition from 0 to 1 as ( x ) moves from ( 2 ) to ( 2 ). This smooth transition ensures that the function is differentiable.

Finally, define ( g(x) ) as the sum of ( h(x) ) and ( k(x) ): [ g(x) = h(x) + k(x) ]

This function ( g(x) ) will be infinitely differentiable and onetoone on ( \mathbb{R} ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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