Why is the graph of $\frac{1}{x}$ a curve?

Question

I don't understand why the graph of $y=\frac{1}{x}$ is a curve. Consider if $x$ doubles. The subsequent value of $y$ would half. When $x$ were to triple, the value of $y$ would divide by $3$.

It seems to me that the relationship is linear. However, when you plot this it is most obviously a curve. What's going on?

Answer 1

This is because, $y$ varies with $x$ inversely; from this relation it gets clear that in the extreme case scenario, asymptotes will be achieved: as $y$ increases manifold, $x$ decreases the same folds and vice-versa. But as this variation is linear, there is a somewhat ‘equilibrium’ point and the approach of the function towards this point gives the hyperbola (a curve) its characteristic shape, except the asymptotes.

Answer 2

The relationship would be linear if $y = c \, x$, where $c$ is a constant coefficient.

But here there is no such constant dependence $y(x)$.

$y_1 - y_2$ varies differently with any $x_1$, $x_2$ you may choose (another way to say it, the derivative $\dfrac{dy}{dx} \neq c$ is not constant).

Hence the relationship between $y$ and $x$ is nonlinear (in this case, a hyperbola).

Answer 3

Here is another way to look at it. If $y = \dfrac{1}{x}$ is a line, then the points $x = 1, 2,3$ would also be equally spaced along the y coordinates because 1, 2, and 3 forms an arithmetic sequence. However, plugging this into $y = \dfrac{1}{x}$, we get the y coordinates are $1, \dfrac{1}{2}$, and $\dfrac{1}{3}$, respectively. This does not form an arithmetic sequence because $2 \cdot \dfrac{1}{2} \neq 1 + \dfrac{1}{3}$. Therefore $y = \dfrac{1}{x}$ cannot be a line; therefore it is a "curve".

Answer 4

Are you perhaps mixing up inverse proportionality $(xy=k)$ and direct proportionality $(\dfrac yx=k)$?

What you are describing (tripling $x$ correspondingly divides $y$ by three) is the former, whilst the latter is actual linearity (tripling $y$ correspondingly triples $x$).

Answer 5

IMHO, one of the most convincing facts is by using a discrete analog : the primary school multiplication table of numbers $1$ to $10$. Spotting equal values and joining them by broken lines suggest th eexistence of hyperbolas with equations $y=\frac{a}{x}$ in the background.

Let us take the example where we join the four occurrences

$$(x,y)=(2,6),(3,4),(4,3),(6,2).$$

where the product

$$xy=12 \tag{1},$$

otherwise said, with :

$$y=\frac{12}{x}$$

It is a broken line because we have taken points with integer coordinates, but if we enlarge the choice of points like (x,y)=(8 ; 1.5)$, they belong also to this curve which is in fact continuous and "smooth"...

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Edit : your question mentions a quest for linearity : one can have linearity when working "at the level of exponents" ; let us take an example : let us consider curve with equation $y=\frac{64}{x}$ ; if we have

$$16=\frac{64}{4} \iff 2^Y=\frac{64}{2^X} \ \ \text{with} \ \ Y=4,X=2,$$

then increasing $Y$ by one unit and decreasing $X$ by one unit still gives $2^Y=\frac{64}{2^X}$, attesting indeed a linear behavior.

The previous considerations amounts to take a log-scale (here base 2 logarithm) as in a previous answer of mine here transforming in fact these hyperbolas into straight lines. Please have a look at the document of Tournès I mention there : it is an exceptionaly didactic presentation of some graphical "computational devices" that have been invented during the last centuries.