# Inverse trig function

## Homework Statement

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

## 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 $\arcsin(x) = y$ or is this the wrong way of solving this?

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Mark44
Mentor

## Homework Statement

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

## 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 $\arcsin(x) = y$ 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:
$f(x) = 3x + 1 + \sin(x)$ with domain $[-\pi/2, \pi/2]$. Without your calculator, determine the value of $f^{-1}(1)$.

Since this is an inverse function, I was going to try to solve for the inverse function, then solve for $f^{-1}(1)$

SammyS
Staff Emeritus
Homework Helper
Gold Member
How is this question related to a problem which requires you to:
Solve $f(x)=1\text{ for }x\,.$​
?

How is this question related to a problem which requires you to:
Solve $f(x)=1\text{ for }x\,.$​
?
To which question are you referring? I never mentioned having to solve $f(x)=1$

ehild
Homework Helper
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
Staff Emeritus
Homework Helper
Gold Member
How is this question related to a problem which requires you to:
Solve $f(x)=1\text{ for }x\,.$​
?
To which question are you referring? I never mentioned having to solve $f(x)=1$
I'll state it more clearly.

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

Solve $f(x)=1\text{ for }x\,.$​

NascentOxygen
Staff Emeritus
The exact wording for this problem:
$f(x) = 3x + 1 + \sin(x)$ with domain $[-\pi/2, \pi/2]$. Without your calculator, determine the value of $f^{-1}(1)$.

Since this is an inverse function, I was going to try to solve for the inverse function, then solve for $f^{-1}(1)$
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

$3x + 1 + \sin(x) = 1$

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

I'll state it more clearly.

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

Solve $f(x)=1\text{ for }x\,.$​
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

$3x + 1 + \sin(x) = 1$

Play around with that equation to see whether you can knock it into something that speaks meaningfully to you.
Ah, I see what you guys are talking about now. Thanks for the help. Much appreciated.