Answer to Question #281828 in Discrete Mathematics for Pooja

Question

15.Determine whether f: R to R, defined as f (x) = −3x + 4 is a bijection. Is f invertible, and if it is, 

what is its inverse?

16.Find the inverse of f (x) =

𝑥+1

𝑥+2

, on a suitable subset of R.

Answer 1

15.

Let $x_1, x_2 \in \R$ and $f(x) = -3x+4.$

If $f(x_1)=f(x_2)=>-3x_1+4=-3x_2+4$

$=>-3x_1=-3x_2=>x_1=x_2.$

If $x_1\not=x_2=>f(x_1)\not=f(x_2).$

Then

$$f(x_1)=f(x_2)<=>x_1=x_2, x_1, x_2\in \R$$

The function $f(x)=-3x+4$ is one-to-one.

$$\forall y\in\R, y=-3x+4, \exist x=-\dfrac{1}{3}y+\dfrac{4}{3}, x\in \R$$

The function $f(x)=-3x+4$ is onto.

The function $f(x)=-3x+4$ is one-to-one and is onto.

Therefore the function $f(x)=-3x+4$ is a bijection. Therefore the function $f(x)=-3x+4$ is invertibele

$$f(x)=-3x+4$$

Replace $f(x)$ with $y$

$$y=-3x+4$$

Switch $x$ and $y$

$$x=-3y+4$$

Solve for $y$

$$y=-\dfrac{1}{3}x+\dfrac{4}{3}$$

Replace $y$ with $f^{-1}(x)$

$$f^{-1}(x)=-\dfrac{1}{3}x+\dfrac{4}{3}$$

16.

$$f(x)=\dfrac{x+1}{x+2}$$

Domain: $(-\infin, -2)\cup (-2, \infin)$

Replace $f(x)$ with $y$

$$y=\dfrac{x+1}{x+2}$$

Switch $x$ and $y$

$$x=\dfrac{y+1}{y+2}$$

Solve for $y$

$$xy+2x=y+1$$

$$y=-\dfrac{2x-1}{x-1}$$

Replace $y$ with $f^{-1}(x)$

$$f^{-1}(x)=-\dfrac{2x-1}{x-1}$$