Dgdrxrt

I am refreshing my high school maths and got an exercise to proof that the graph of a linear function and its inverse cannot be perpendicular. Below is my proof.

1. A linear function is a straight line.


2. An inverse function is a reflection of the function of $x = y$ line. This means the function and its inverse must have the same angle when it meet, which is 45.


3. From 2., $f(x)$ must be parallel with y axis and $f^-1(x)$ must be parallel with x axis (or vise-versa.)


4. If $f(x)$ is parallel with $y$ axis, then it's not a function.

Is my proof correct? Are there any other ways of proving this?

The proof you give is more or less a sketch. It uses some not well-defined notions. A mathematical proof could run like this:

The linear function $f(x) = k x + d$ has direction vector $(1,k)$. The inverse $f^-1(x) = fracx-dk$ only exists for $k ne 0$ and has direction vector $(1,frac1k)$.

For the scalar product of the two direction vectors we get:

$(1,k) cdot (1,frac1k) = 1 + k frac1k = 1 + 1 = 2$

which is non-zero, therefore the graphs are not perpendicular.

You are right. For clarity, you should explain were the $45°$ is coming from. Also beware that you should discuss the case of negative coefficients.

If $f$ is a linear function, then $f(x) = ax + b$, with $a neq 0$. The graph of $f$ has slope $a$. Since $a neq 0$, the graph of $f$ is not horizontal. The graph of $f$ is not vertical since a function of $x$ has a unique $y$ value for each $x$ value.

We solve the inverse by interchanging $x$ and $y$, then solving for $y$.
beginalign*
y & = ax + b\
x & = ay + b && textinterchange variables\
x - b & = ay\
fracx - ba & = y && textdivision by $a neq 0$ is defined\
frac1ax - fracba & = y
endalign*
Hence, the inverse of $f$ is
$$g(x) = frac1ax - fracba$$
which you can verify by showing that $(g circ f)(x) = x$ and $(f circ g)(x) = x$ for each real number $x$. The product of the slopes of the graphs of $f$ and $g$ is
$$a cdot frac1a = 1$$
However, the product of the slopes of non-vertical perpendicular lines is $-1$. Hence, the graph of a linear function and its inverse cannot be perpendicular.

Note that if $a = 0$, then $f(x) = b$ is a constant function rather than a linear function. A constant function does not have an inverse since there is more than one value of $x$ for each value of $y$. If a function does have an inverse, then a horizontal line will cross its graph at most once (so we can write $x$ as a function of $y$), which is known as the Horizontal Line Test. A constant function fails the Horizontal Line Test.