How can I prove this function is bijective?

Question

I know I have to show it's injective and surjective, but up until now it's always been simple equations that I can prove are equal to each other. This equation is a little bit more complex.

$$f(x) = (x-a)\frac{d-c}{b-a} + c$$

Im $99\%$ sure it's bijective, but I don't know how to prove it since there is nothing to set it equal to like other examples. Also $a \lt b$ and $c \lt d$

Answer 1

I assume you operate in the set $\mathbb{R}$.

Let's use the abbreviation $m = \frac{d-c}{b-a}$ $$\color{blue}{f(x)} = m(x-a)+c = mx +c-ma \color{blue}{= mx + r} \mbox{ with } r = c-ma \mbox{ and } m>0$$ Injectivity:

$f$ is strictly increasing, hence it is injective: $$x<y \Rightarrow mx+r < my+r \Rightarrow f(x) < f(y)$$ Surjectivity:

For each $y \in \mathbb{R}$ you need to find an $x \in \mathbb{R}$ such that $f(x) = y$: $$y = mx+r \Rightarrow x=\frac{y-r}{m} \mbox{ is such an }x$$

Answer 2

Clearly, the function is continuous assuming the domain being $[a,b]$ and for $x=a\implies f(a)=c$ and $x=b\implies f(b)=d$ again for $x=\frac{a+b}{2}$ we ave $f(\frac{a+b}{2})=\frac{c+d}{2}$. So $f$ assumes every value in $[a,b]$ uniquely in $[c,d]$. Hence it is a bijection. Hope this makes the $1\%$ complete.

Answer 3

Assuming the range and domain are $\mathbb{R}$ (or any field), it's bijective, as is any non-constant linear function; if $$ f(x) = mx + b $$ where $m \neq 0$, then $$ f^{-1}(x) = \frac{1}{m}(x - b). $$