# One-to-One Intervals Problem #5 - UGNotesOnline

• lutya2
In summary, the conversation discusses problem #5 on a given worksheet and the attempts at solving it. Part (a) involves attempting to take the derivative of a function in the integral to prove its one-to-one property, while part (b) asks for the evaluation of a derivative at a specific point. The conversation also provides hints and reminders for solving the problem.
lutya2

## Homework Statement

I'm trying to do problem #5 on this worksheet
http://ugnotesonline.com/attachments/008_t1.5pst.pdf

none

## The Attempt at a Solution

(a) I tried to take the derivitive of the function in the intergral so I could see where it was increasing and decreasing and prove that it was one-to-one but that didn't work.

(b) I don't even know what this part is trying to ask.

Last edited by a moderator:
lutya2 said:
(a) I tried to take the derivitive of the function in the intergral so I could see where it was increasing and decreasing and prove that it was one-to-one but that didn't work.
Try again and show us your work because that is as far as I can see the easiest approach. Remember that if f is everywhere differentiable and the derivative is strictly positive, then the function is strictly increasing.

For b, the notation $f^{-1}(x)$ denotes the inverse function. That is the function such that
$$f^{-1}(f(x)) = f(f^{-1}(x)) = x$$
for all x. (its existence is guaranteed by part a). To solve this part use the chain rule as follows:
$$\frac{d(f \circ f^{-1})(x)}{df^{-1}} \times \frac{df^{-1}(x)}{dx} = \frac{d(f^{-1} \circ f)(x)}{dx}$$
You should be able to evaluate everything here except $$\frac{df^{-1}(x)}{dx}$$ so you can solve the problem.

HINT:
$$\frac{d(f \circ f^{-1})(x)}{df^{-1}} = \frac{df(x)}{dx}$$

EDIT:
Also in case you didn't know $$g|_{x=0}$$ just means g(0), so in b you're just asked to evaluate the derivate at 0.

## 1. What is the purpose of "One-to-One Intervals Problem #5 - UGNotesOnline"?

The purpose of this problem is to practice finding one-to-one intervals for a given function and to understand the concept of one-to-one intervals and their relationship to the graph of a function.

## 2. How do I solve "One-to-One Intervals Problem #5 - UGNotesOnline"?

To solve this problem, you will need to first determine the intervals where the function is increasing or decreasing. Then, you will need to check for horizontal and vertical shifts and reflections. Finally, you will need to determine the one-to-one intervals by analyzing the graph of the function.

## 3. What is a one-to-one interval?

A one-to-one interval is a portion of the domain of a function where the function is both increasing or decreasing and one-to-one, meaning that each input has a unique output.

## 4. Why is understanding one-to-one intervals important?

Understanding one-to-one intervals is important because it helps us to determine which parts of the graph of a function are one-to-one and which are not. This can be useful in finding inverse functions and understanding the behavior of functions.

## 5. Can I use a calculator to solve "One-to-One Intervals Problem #5 - UGNotesOnline"?

While you may use a calculator to check your work, it is recommended to solve this problem by hand. This will help you to better understand the concept of one-to-one intervals and how they relate to the graph of a function.

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