First Principles AI

Solution

Let:

f(x)=\frac{2x-3}{x+1}

Domain: denominator $x+1\ne 0\Rightarrow x\ne -1$ .

\boxed{D_f=\mathbb{R}\setminus\{-1\}}

For range, let $y=\dfrac{2x-3}{x+1}$ and solve for $x$ :

\begin{aligned} y(x+1)&=2x-3\\ yx+y&=2x-3\\ yx-2x&=-3-y\\ x(y-2)&=-(3+y)\\ x&=\frac{y+3}{2-y} \end{aligned}

So $2-y\ne 0\Rightarrow y\ne 2$ .

\boxed{R_f=\mathbb{R}\setminus\{2\}}

Co-domain is $\mathbb{R}$ but range is $\mathbb{R}\setminus\{2\}$ , so $f$ is not onto.

\boxed{f\ \text{is not onto}}

Assume $f(x_1)=f(x_2)$ :

\frac{2x_1-3}{x_1+1}=\frac{2x_2-3}{x_2+1}

Cross-multiply:

(2x_1-3)(x_2+1)=(2x_2-3)(x_1+1)

Expand:

2x_1x_2+2x_1-3x_2-3=2x_1x_2+2x_2-3x_1-3

Cancel $2x_1x_2$ and $-3$ :

2x_1-3x_2=2x_2-3x_1\Rightarrow 5x_1=5x_2\Rightarrow x_1=x_2

So $f$ is one-to-one.

\boxed{f\ \text{is one-to-one}}