Do all straight lines have an inverse function? It would seem to make sense. All linear lines would pass the horizontal line test and thus when reflected across $y=x$ it would still be a function. However, the answers says no, and I cannot find a case where my logic doesn't work. Could I have some insight please.
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$\begingroup$Characteristic for non-vertical "straight lines" is that they correspond with functions that can be prescribed by $x\mapsto ax+b$ where $a,b$ are fixed real numbers.
Based on equation $y=ax+b$ we can find an expression for $x$ in $y$ under the extra condition that $a\neq0$: $$x=\frac1{a}(y-b)$$By interchanging $x$ and $y$ we find the inverse function is:$$y=\frac1{a}(x-b)$$ This tells us that such linear functions have an inverse if $a\neq0$. In case $a=0$ we are dealing with a constant function prescribed by $x\mapsto b$. This function is not injective hence has no inverse.
$\endgroup$ $\begingroup$1) Answer is no:
$X- $axis: $y = f(x) = 0$, $ x \in \mathbb{R} $, and
$Y-$axis : $x= g(y) = 0$, $y \in \mathbb{R}$
do not have inverses.
2) Let $y = f(x) = mx + c$, with $m \ne 0$, $m$ real.
Then $y' = f'(x) = m $;
A) $m\gt 0$: $f'(x) = m \gt 0$, i.e.
$f(x)$ is strictly mon. increasing everywhere.
B) $m\lt 0$: $f'(x) = m \lt 0$, i.e.
$f(x)$ is strictly mon. decreasing everywhere.
$f(x)$ in case A) or B) has an inverse functions.
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