Chapter 5.2, Problem 72AYU

In Problems 63-74, the function $f$ is one-to-one. (a) Find its inverse function $f^{-1}$ and check your answer. (b) Find the domain and the range of $f$ and $f^{-1}$.

$f(x)=\frac{x^2+3}{3x^2}$ $x>0$

To determine

To find:

a. Find its inverse function $f^{-}{}^1$ and check your answer.

$f(x)=\frac{x^2+3}{3x^2}$, $x>0$

Answer to Problem 72AYU

Solution:

a. $f^{-1}\left(x\right)=\frac{\sqrt{3}}{\sqrt{3x-1}},x>0$

Explanation of Solution

Given:

i.e, A function $f$ is one-to-one if for all $a,b\in A$ and $a\ne b\Rightarrow f(a)\ne f(b)\Rightarrow x\ne y$.

Now, $f\left(a\right)=x\Rightarrow f^{-1}\left(f\left(a\right)\right)=f^{-1}\left(x\right)\Rightarrow a=f^{-1}\left(x\right)$.

Calculation:

a. Let $f\left(x\right)=\frac{x^2+3}{3x^2}=y$.

$\Rightarrow$ i.e., $\frac{x^2+3}{3x^2}=y$

$\Rightarrow x^2+3=y\left(3x^2\right)$

$\Rightarrow x{}^2+3=3x^{2}y$

$\Rightarrow 3x^{2}y-x^2=3$

$\Rightarrow x^2\left(3y-1\right)=3$

$\Rightarrow x^2=\frac{3}{3y-1}$

$\Rightarrow$ Change $x$ to $y$ and $y$ to $x$.

$\Rightarrow y^2=\frac{3}{3x-1}\Rightarrow y=\pm \frac{\sqrt{3}}{\sqrt{3x-1}}$

$\Rightarrow$ Rewriting this, we get

$f^{-1}\left(x\right)=\frac{\sqrt{3}}{\sqrt{3x-1}},x>0$

Now,

$\left(f.f^{-1}\right)\left(x\right)=f(f^{-1}\left(x\right))$

$\left(f.f^{-1}\right)\left(x\right)=f\left(\frac{\sqrt{3}}{\sqrt{3x-1}}\right)$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{{(\frac{\sqrt{3}}{\sqrt{3x-1}})}^2+3}{3{(\frac{\sqrt{3}}{\sqrt{3x-1}})}^2}$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{\frac{3}{3x-1}+3}{3\frac{3}{3x-1}}$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{\frac{3+9x-3}{3x-1}}{\frac{9}{3x-1}}$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=\frac{9x}{9}$

$\Rightarrow \left(f.f^{-1}\right)\left(x\right)=x$

Also,

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=f^{-1}\left(f\left(x\right)\right)$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=f^{-1}\left(\frac{x^2+3}{3x^2}\right)$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{\sqrt{3}}{\sqrt{3\left(\frac{x^2+3}{3x^2}\right))-1}}$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{\sqrt{3}}{\sqrt{\frac{x^2+3-x^2}{x^2}}}$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=\frac{\sqrt{3}x}{\sqrt{3}}$

$\Rightarrow \left(f^{-1}.f\right)\left(x\right)=x$

$\Rightarrow$ Here we verified that $\left(f.f^{-1}\right)\left(x\right)=x=\left(f^{-1}.f\right)\left(x\right)$.

To determine

To find:

b. Find the domain and the range of $f$ and $f^{-}{}^1$.

$f(x)=\frac{x^2+3}{3x^2}$, $x>0$

Answer to Problem 72AYU

Solution:

b. The domain of $f\text{ = Range of }{f}^{\text{–1}}\text{ = }\left\{x\text{|}x>0\right\}$.

The range of $f\text{ = Domain of }{f}^{\text{–1}}\text{ = }\left\{x\text{|}x>\frac{1}{3}\right\}$.

Explanation of Solution

Given:

i.e, A function $f$ is one-to-one if for all $a,b\in A$ and $a\ne b\Rightarrow f(a)\ne f(b)\Rightarrow x\ne y$.

Now, $f\left(a\right)=x\Rightarrow f^{-1}\left(f\left(a\right)\right)=f^{-1}\left(x\right)\Rightarrow a=f^{-1}\left(x\right)$.

Calculation:

b. When working with rational functions, domain elements must not create division by zero. Any $x$ not in the domain of $f$ must be excluded. So, we know that the domain of $f(x)$ cannot contain 0.

For this, $f\left(x\right)=\frac{x^2+3}{3x^2}$ shows us that $x=0$ would not be an acceptable domain element, since it creates a zero denominator, setting $x=0$. Also exclude 0 from the domain of $f(x)$. The domain of $f(x)$ is $\left\{x\text{|}x>0\right\}$.

For this, $f^{-1}\left(x\right)=\frac{\sqrt{3}}{\sqrt{3x-1}}$ shows us that $x=\frac{1}{3}$ would not be an acceptable domain element, since it creates a zero denominator, setting $3x-1=0,3x-1>0.3x>1,x>\frac{1}{3}$.

Also exclude $\frac{1}{3}$ from the domain of $f^{-1}\left(x\right)$.

The domain of $f^{-1}\left(x\right)$ is $\left\{x\text{|}x>\frac{1}{3}\right\}$.

Therefore, The domain of $f\text{ = Range of }{f}^{\text{–1}}\text{ = }\left\{x\text{|}x>0\right\}$.

The range of $f\text{ = Domain of }{f}^{\text{–1}}\text{ = }\left\{x\text{|}x>\frac{1}{3}\right\}$.