To find the range of a given ##\sin## function

[brotherbobby]

Homework Statement

Find the $\textbf{range}$ of the following function : $\boldsymbol{f(x) = \sin(3x^2+1)}$

Relevant Equations

The range of the $\sin x$ curve is between the limits of $-1$ and $+1$, for $x\in (-\infty, +\infty)$

Attempt : The domain of the function $\sin(3x^2+1)$ is clearly $x\in (-\infty, +\infty)$. The values of $x$ go into all quadrants where the $\sin$ curve is positive and negative. Hence the range of the function : $\boxed{\mathscr{R}\{f(x)\}=(-1,+1)}\quad\color{green}{\Large{\checkmark}}$.

Doubt : This is fine and agrees with the answer in the text. However there is a different way in which I evaluate the range of a function, especially when it is not obvious. I invert $y=f(x)$ to express $x = f^{-1}(y)$ and then go on to find the domain of the inverse function. That gives me the acceptable values of $y$, the range of the original function.

So let me do it here : $\small{y = \sin{3x^2+1}\Rightarrow 3x^2+1=\sin^{-1}y\Rightarrow 3x^2=\sin^{-1}y-1\Rightarrow \boldsymbol{x = \sqrt{\dfrac{\sin^{-1}y-1}{3}}}}$

What are the acceptable values of $y$? We must have $\sin^{-1}y-1\ge 0$, since it is under root. From here I obtain that the only acceptable value for $y$ is $\boxed{+1}$, which is the range of the original function $f(x)$.

Request : Can someone tell me where is my mistake in the second method? Many thanks.

 

[FactChecker]

I invert $y=f(x)$ to express $x = f^{-1}(y)$

Analytically inverting a function is often an extremely difficult thing.

and then go on to find the domain of the inverse function.

Ok. But the domain of the inverse is the same as the range of the original, so this is just rewording the problem.

That gives me the acceptable values of $y$, the range of the original function.

So let me do it here : $\small{y = \sin{3x^2+1}\Rightarrow 3x^2+1=\sin^{-1}y\Rightarrow 3x^2=\sin^{-1}y-1\Rightarrow \boldsymbol{x = \sqrt{\dfrac{\sin^{-1}y-1}{3}}}}$

What are the acceptable values of $y$?

Doesn't this depend on the range of the original function? Anywhere else, the function is not even defined.
If the original function was defined only for x in [1/4, 1.2], then you would still need to determine the matching values of y.

 

[brotherbobby]

Doesn't this depend on the range of the original function? Anywhere else, the function is not even defined.
If the original function was defined only for x in [1/4, 1.2], then you would still need to determine the matching values of y.

I didn't understand this bit at all.

 

[FactChecker]

I didn't understand this bit at all.

This problem has something that makes it easier than others might be. The original function is defined (we assume) for all $x \in (-\infty. \infty)$ so you don't have to worry about any restrictions on $x$. What if the problem had said: Find the range of $f(x) = \sin( {3x^2+1})$ for $x \in (1/4, 1/2)$? Then you would have to find the range of the function that was associated with $x \in (1/4, 1/2)$, which is harder.

 

[Hill]

Homework Statement: Find the $\textbf{range}$ of the following function : $\boldsymbol{f(x) = \sin(3x^2+1)}$
Relevant Equations: The range of the $\sin x$ curve is between the limits of $-1$ and $+1$, for $x\in (-\infty, +\infty)$

Can someone tell me where is my mistake in the second method?

$sin^{-1}$ is a multivalued function and thus the derivation in the second method should be rather, for integer $n$: $$\small{y = \sin{(3x^2+1)}\Rightarrow 3x^2+1+2\pi n=\sin^{-1}y\Rightarrow 3x^2=\sin^{-1}y-1-2\pi n\Rightarrow \boldsymbol{x = \pm \sqrt{\dfrac{\sin^{-1}y-1-2\pi n}{3}}}}$$
(the square root function is also multivalued).
This allows for any $y$.

 

[PeroK]

First of all, formally the range of a composite function $f \circ g$ is the range of $f$ restricted to the range of $g$. In this case, the range of $3x^2 + 1$ is $[1, \infty)$ and the range of $\sin x$ restricted to $[1, \infty)$ is $[-1, 1]$. You don't need any further calculation for a well-known function such as $\sin x$.

If you wanted to do more calculations in this case, then something like this is needed. First note that the range of $\sin(3x^2 +1)$ is a subset of $[-1, 1]$, which is the range of $\sin x$.

Let $z \in [-1, 1]$. We can find $\theta \in [0, 2\pi]$ such that $\sin \theta = z$. Let $y = \theta + 2\pi$. Then $y > 1$ and $\sin y = \sin \theta = z$. Finally, let $x = \sqrt{\dfrac{y - 1}{3}}$ (so that $3x^2 + 1 = y$). Then we have found $x$ such that $\sin(3x^2 + 1) = z$. QED

 

[PeroK]

$sin^{-1}$ is a multivalued function and thus the derivation in the second method should be rather, for integer $n$: $$\small{y = \sin{(3x^2+1)}\Rightarrow 3x^2+1+2\pi n=\sin^{-1}y\Rightarrow 3x^2=\sin^{-1}y-1-2\pi n\Rightarrow \boldsymbol{x = \pm \sqrt{\dfrac{\sin^{-1}y-1-2\pi n}{3}}}}$$
(the square root function is also multivalued).
This allows for any $y$.

I'm not convinced this is a sound argument!

 

[Hill]

I'm not convinced this is a sound argument!

The OP has asked to find a mistake in their derivation. I argue that mistake is here:

From here I obtain that the only acceptable value for y is +1,

The mistake is to consider one value only of the function $sin^{-1}$. For example, the OP assumes that $y=0$ is not acceptable because it makes $\sqrt {-1}$ in their last expression. However, it is in fact acceptable, because $sin^{-1} (0)$ is not only $0$, but any $\pi n$. Similarly for all other values of $y \in [-1, +1]$.

 

[PeroK]

The OP has asked to find a mistake in their derivation. I argue that mistake is here:

The mistake is to consider one value only of the function $sin^{-1}$. For example, the OP assumes that $y=0$ is not acceptable because it makes $\sqrt {-1}$ in their last expression. However, it is in fact acceptable, because $sin^{-1} (0)$ is not only $0$, but any $\pi n$. Similarly for all other values of $y \in [-1, +1]$.

It's better than the OP and it's part of a solution, but the argument doesn't stand up formally. You really need to start with $x$ and explicitly cover the intermediate range. I gave an informal and a formal solution in post #6.

It is "precalculus", so I think looking at the graph of $\sin x$ is a better approach. If that's not good enough, then you need to do things properly and formally.

PS technically $\sin^{-1}$ is single-valued function.

 

[Hill]

BTW, the answer should be not $\mathscr{R}\{f(x)\}=(-1,+1)$, but rather $\mathscr{R}\{f(x)\}=[-1,+1]$.

 

[Hill]

PS technically sin−1 is single-valued function.

Yes, that's why I've added $2 \pi n$ in my derivation in #5. (And the $\pm$ in front of the square root.)

 

[PeroK]

Yes, that's why I've added $2 \pi n$ in my derivation in #7. (And the $\pm$ in front of the square root.)

I guess posts #5 and #6 represent the different approaches of a physics and maths major, respectively!

 

[Hill]

I guess posts #5 and #6 represent the different approaches of a physics and maths major, respectively!

Applied math, actually.

 

[Gavran]

The function $$ f : ( - \infty , + \infty ) \to ( - 1 , + 1 ) $$ where $$ f ( x ) = \sin ( 3 x ^ 2 + 1 ) $$ is not a bijective function and the inverse function $$ f ^ { - 1 } ( x ) $$ does not exist. This is the reason why the second method is not applicable.

 

[WWGD]

The function $$ f : ( - \infty , + \infty ) \to ( - 1 , + 1 ) $$ where $$ f ( x ) = \sin ( 3 x ^ 2 + 1 ) $$ is not a bijective function and the inverse function $$ f ^ { - 1 } ( x ) $$ does not exist. This is the reason why the second method is not applicable.

Still, I believe local inverses exist. If when f'(x) is none zero.

 

[Gavran]

Of course, local inverses for the function $$ f ( x ) = \sin ( 3 x ^ 2 + 1 ) $$ exist. For example, the function $$ f : \left [ 0 , \sqrt { \frac { \pi - 2 } { 6 } } \right ] \to \left [ \sin 1 , 1 \right ] $$ where $$ f ( x ) = \sin ( 3 x ^ 2 + 1 ) $$ is a bijective function.