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## Here's the Solution to this Question

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 $f^{-1}(x)$

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