Solve Inverse Trig Function with Domain [-π/2, π/2] | Step-by-Step Guide

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Homework Help Overview

The problem involves finding the inverse of the function f(x) = 3x + 1 + sin(x) within the domain [-π/2, π/2]. The specific task is to determine the value of f^{-1}(1) without using a calculator.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of setting y = f(x) and the implications of finding the inverse function. There is uncertainty about whether the problem requires solving for the inverse directly or finding the x value that satisfies f(x) = 1.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have suggested that the equation 3x + 1 + sin(x) = 1 should be manipulated to find x, while others are questioning the relationship between finding the inverse and solving f(x) = 1.

Contextual Notes

There is some confusion regarding the exact requirements of the problem, particularly whether it is necessary to find the inverse function or simply solve for the x value that results in f(x) = 1. The domain of the function is also a critical factor in the discussion.

togame
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Homework Statement


My problem is as follows:
find the inverse of
[tex]3x+1+\sin(x)[/tex] with the domain [itex][-\frac{\pi}{2},\frac{\pi}{2}][/itex]


Homework Equations





The Attempt at a Solution


for this would I just try to solve as normal by setting y=f(x) then using the fact that [itex]\arcsin(x) = y[/itex] or is this the wrong way of solving this?
 
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togame said:

Homework Statement


My problem is as follows:
find the inverse of
[tex]3x+1+\sin(x)[/tex] with the domain [itex][-\frac{\pi}{2},\frac{\pi}{2}][/itex]


Homework Equations





The Attempt at a Solution


for this would I just try to solve as normal by setting y=f(x) then using the fact that [itex]\arcsin(x) = y[/itex] or is this the wrong way of solving this?

You're not going to be able to solve the equation y = f(x) = 3x + 1 + sin(x) for x (to get the inverse x = f-1(y).
What is the exact problem statement? It might be that you are misreading what is being asked for in this problem.
 
The exact wording for this problem:
[itex]f(x) = 3x + 1 + \sin(x)[/itex] with domain [itex][-\pi/2, \pi/2][/itex]. Without your calculator, determine the value of [itex]f^{-1}(1)[/itex].

Since this is an inverse function, I was going to try to solve for the inverse function, then solve for [itex]f^{-1}(1)[/itex]
 
How is this question related to a problem which requires you to:
Solve [itex]f(x)=1\text{ for }x\,.[/itex]​
?
 
SammyS said:
How is this question related to a problem which requires you to:
Solve [itex]f(x)=1\text{ for }x\,.[/itex]​
?

To which question are you referring? I never mentioned having to solve [itex]f(x)=1[/itex]
 
The problem asks to determine f-1(1)=x. The equation is equivalent to 1 =f(x), that is, 3x+1+sin(x)=1. Solve for x.

ehild
 
SammyS said:
How is this question related to a problem which requires you to:
Solve [itex]f(x)=1\text{ for }x\,.[/itex]​
?

togame said:
To which question are you referring? I never mentioned having to solve [itex]f(x)=1[/itex]
I'll state it more clearly.

How are the following two problems related?
Determine the value of [itex]f^{-1}(1)\,.[/itex]

Solve [itex]f(x)=1\text{ for }x\,.[/itex]​
 
togame said:
The exact wording for this problem:
[itex]f(x) = 3x + 1 + \sin(x)[/itex] with domain [itex][-\pi/2, \pi/2][/itex]. Without your calculator, determine the value of [itex]f^{-1}(1)[/itex].

Since this is an inverse function, I was going to try to solve for the inverse function, then solve for [itex]f^{-1}(1)[/itex]
While that is the general method, if you are lucky the problem is one that can be solved by recognizing it to be a special case that is easy to solve without ploughing all the way through the full general method (even if the general method were possible).

So you are being asked to find the x value (or values) that makes

[itex]3x + 1 + \sin(x) = 1[/itex]

Play around with that equation to see whether you can knock it into something that speaks meaningfully to you. :smile:
 
SammyS said:
I'll state it more clearly.

How are the following two problems related?
Determine the value of [itex]f^{-1}(1)\,.[/itex]

Solve [itex]f(x)=1\text{ for }x\,.[/itex]​

NascentOxygen said:
While that is the general method, if you are lucky the problem is one that can be solved by recognizing it to be a special case that is easy to solve without ploughing all the way through the full general method (even if the general method were possible).

So you are being asked to find the x value (or values) that makes

[itex]3x + 1 + \sin(x) = 1[/itex]

Play around with that equation to see whether you can knock it into something that speaks meaningfully to you. :smile:

Ah, I see what you guys are talking about now. Thanks for the help. Much appreciated.
 

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