# Inverse functions, and limits help

• johnq2k7
In summary: One way is to show that there is a unique f(x) value that is increasing for all x values in the domain. For example, if you are looking at the function y=x^3+cos x, then the value of f(x) at x=2 is 3+cos 2x, but the value of f(x) at x=3 is -1+cos 3x. So there is a unique f(x) value that is increasing for all x values in the domain. Another way to show that a function is monotonic on its domain is to show that there is a unique decreasing f(x) value for all x values in the domain.
johnq2k7

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

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?

## What are inverse functions?

Inverse functions are functions that "undo" each other. In other words, if a function takes an input and produces an output, its inverse will take that output and produce the original input. To find the inverse of a function, you must switch the x and y variables and solve for y.

## How do you find the inverse of a function?

To find the inverse of a function, you must switch the x and y variables and solve for y. This can be done by using algebraic techniques such as factoring, completing the square, or using the quadratic formula.

## What is the notation for inverse functions?

The notation for inverse functions is f-1(x), where f(x) is the original function. This notation represents the inverse function that "undoes" the original function.

## What is a limit?

A limit is a fundamental concept in calculus that describes the behavior of a function as the input approaches a certain value. It is essentially the value that a function is approaching, rather than the actual value at that point.

## How do you evaluate limits?

To evaluate limits, you can use various techniques such as direct substitution, factoring, and algebraic manipulation. If these techniques do not work, you may need to use more advanced methods such as L'Hopital's rule or the squeeze theorem.

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