Inverse functions, and limits help

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Homework Help Overview

The discussion revolves around the properties of inverse functions and limits, specifically focusing on determining the invertibility of certain functions over specified intervals and evaluating various limits. The subject area includes concepts from calculus and analysis, particularly dealing with functions and their behaviors.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conditions under which the function sech x is invertible on the interval [0, infinity), with some suggesting the need to demonstrate that the function is one-to-one by using the horizontal line test or by showing monotonicity. Questions arise about the proper methods to establish these properties.

Discussion Status

The conversation is ongoing, with participants exploring different approaches to determine the invertibility of the function. Some guidance has been provided regarding the importance of monotonicity and the distinction between finding an inverse and proving its existence.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the use of certain resources or methods. There is an emphasis on justifying answers for limits and understanding the conditions for invertibility without directly providing solutions.

johnq2k7
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Please help me with these following problems:

1.)Indicate whether each of the following functions is invertible in the given interval. Explain
a.) sech x on [0,infinity)

b.) cos (ln x) on (O, e^pie]

c.) e^(x^2) on (-1,2]


2.) Evaluate the following limits, justifying your answers. If a limit does not exist explain why.

a.) lim (x--> inf.) (3x^3 +cos x)/(sin x- x^3)
b.) lim (x-->Pie(+)) (tan^-1 (1/(x-Pie)))/(Pie-x)
c.) lim (x--> 0(+)) (sqrt(x+sin x))(ln x)
d.) lim (x--> 1(-)) (cos^-1(x))/(1-x)

3.) give the equation of the line tangent to the curve at the given point.
a.) (y)(tan^-1 x)= x*y at (sqrt(3),0)
b.) ln y= x^2 +(2)*e^x at (0, e^2)


My attempt at a solution for a.)

1.) sech x on [0,infinity)

let y= sech x
therefore, y*sech^-1 x= sech^-1 x* sech x

therefore x= y*sech^-1 x

since it is an inverse function switch x and y variables

therefore f^-1(x)= x*sech^-1 (y)

how do i determine, it's intervible in [0, infinity)... I'm not sure if this prior work is correct
 
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johnq2k7 said:
1.) sech x on [0,infinity)

let y= sech x
therefore, y*sech^-1 x= sech^-1 x* sech x

therefore x= y*sech^-1 x

since it is an inverse function switch x and y variables

therefore f^-1(x)= x*sech^-1 (y)

That won't do. You've simply assumed the existence of sech^-1(x). You need to show that the function is 1-1 on [0,\infty) (that is, you need to show that it passes the so-called horizontal line test on that interval). One way to do that is to show that the function is monotonic on that interval. Have you been taught a theorem that shows you how to determine the monotonicity of a function?
 
yes, basically for the horizontal line test, each x values only one unique f(x) value if it is to be one-to-one...

since, sech(x) = 1 / cosh(x) = 2 / (e^x + e^-x)

can i compute y= 2/(e^x +e^-x)

and solve for x:

than switch the variables for y and x to find the inverse function?

and then plug in the values of [0,infinity) into x

and determine if a f(x) value exists?

is this a proper method
 
In principle that's fine, but in practice it's a pain. It would be easier to show that y=sech(x) is monotonic (that is, either always increasing or always decreasing). Now the question for you is, how do you tell whether a function is increasing or decreasing on an interval?
 
but wouldn't I have to determine the graph of the inverse of the sech (x) equation in order to determine if a value exists for [0, infinity)
 
You don't have to find the inverse, you just have to determine whether or not an inverse exists. There's a big difference. You need to use the following facts.

1.) A function f:D\longrightarrow R has an inverse f^{-1}:R\longrightarrow D if and only if f is 1-1 and onto. The onto part is easy: just assume that the output values of the function are the range of the function. It's the 1-1 part that you need to worry about. This brings is to...

2.) A function is 1-1 on its domain if and only if it is monotonic on its domain.

So how do you show that f is monotonic?
 

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