How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain.

Question

$f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$

I know I have to switch the $f(x)$ and the $y$:

$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$

To find the domain:

$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$

$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$

$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$

$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$

The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$

But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.

Thanks

Answer 1

Look at the plot below (first picture). The function can be inverted only in the interval where it is bijective.

Take the derivative $$f'(x)=\frac{4}{3} \sin \left(\frac{1}{3} (2 x+1)\right) \cos \left(\frac{1}{3} (2 x+1)\right)=\frac{2}{3} \sin \left(\frac{2}{3} (2 x+1)\right)$$ $$f'(x)=0\to \sin \left(\frac{2}{3} (2 x+1)\right)=0$$ In the interval we are interested in, this happens when $$\frac{2}{3} (2 x+1)=0;\;\frac{2}{3} (2 x+1)=\pi$$ which gives $$x_1=-\frac12;\;x_2=\frac{3 \pi }{4}-\frac{1}{2}$$ We will invert $f(x)$ in the interval $[x_1,x_2]$.

Set $y=f(x)$ and get $$\sin ^2\left(\frac{1}{3} (2 x+1)\right)=y\to \sin \left(\frac{1}{3} (2 x+1)\right)=\sqrt{y}$$ then $$\frac{1}{3} (2 x+1)=\arcsin\sqrt y\to 2x+1=3\arcsin\sqrt y\to x=\frac{1}{2} \left(3 \arcsin\left(\sqrt{y}\right)-1\right)$$ Only now we swap $x$ and $y$ to get $$f^{-1}(x)=\frac{1}{2} \left(3 \arcsin\left(\sqrt{x}\right)-1\right)$$ The domain of $f^{-1}(x)$ is the range of $f(x)$ that is $[0,1]$.

Look at the second picture below. The inverse function graph is the symmetric of the function $f(x)$ wrt the line $y=x$.

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Answer 2

a)
The function $f$ is not injective, therefore not bijective and cannot have an inverse therefore.
But you can calculate the individual branches (i. e. partial inverses) of $f$.

b)
You didn't rearrange your equation correctly.

Let $y$ denote $f^{-1}(x)$.

$$\sin^2\left(\frac{2y+1}{3}\right)=x$$

$$\left(\sin\left(\frac{2y+1}{3}\right)\right)^2=x$$

$$\sin\left(\frac{2y+1}{3}\right)=\pm\sqrt{x}$$

$$\frac{2y+1}{3}=\arcsin(\pm\sqrt{x})$$

$$\frac{2y+1}{3}=\pm\arcsin(\sqrt{x})$$

$$y=\pm\frac{3}{2}\arcsin(\sqrt{x})-\frac{1}{2}$$

Answer 3

$$\text{Let $y^2 = f(x)=\sin^2\left(\frac{2x+1}{3}\right)$}$$

So $y \in [-1,1]$ and $y=\sin\left(\dfrac{2x+1}{3}\right)$

where $\dfrac{2x+1}{3} = 2n\pi + \theta $ and $-\pi \le \theta \le \pi$ for some $n \in \mathbb Z$.

We compute

\begin{align} \theta &= \arcsin y \\ \dfrac{2x+1}{3} &= 2n \pi + \arcsin y \\ 2x &= 6n \pi - 1 + 3 \arcsin y \\ x &= 3n \pi - \dfrac 12 + \dfrac 32 \arcsin y \\ \end{align}

Answer 4

As IV_ said, your inverse function should be $y=\pm\frac{3}{2}\arcsin(\sqrt{x})-\frac{1}{2}$.

Your original function $ f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ is equal to a sine function squared. If you graph f(x), you can see that its biggest value is $1$, and it achieves its smallest value at $x=(-3\pi+1)/2$ because the domain is restricted. Calculating the smallest $y$-value, we get $f((-3\pi+1)/2)$, which is approximately $0.382$ according to the graph. Thus, the range of $f(x)$ is $[0.382,1]$.

So, your function maps $x$-values in its domain to the range $[0.382,1]$. An inverse function reverses this process, so it maps $[0.382,1]$ back to the domain of the original function. That means that the domain of your inverse function is $[0.382,1]$.

Notice how we didn't have to find the domain of the original function to find the domain of the inverse?