y = x + \dfrac{1}{x}

Graph of the Rational Function $y = x + \dfrac{1}{x}$

$y = x + \dfrac{1}{x}$ is just a linear function added to a reciprocal, yet it is the standard example of a curve with a slant asymptote, and it is the function in which the inequality between the arithmetic and geometric means becomes visible as a shape.

Domain and symmetry

The domain is $x \neq 0$ . Since $f(-x) = -f(x)$ , the function is odd and the graph is symmetric about the origin.

Monotonicity and extrema

The derivative is $f'(x) = 1 - \dfrac{1}{x^2} = \dfrac{x^2-1}{x^2}$ . The denominator is always positive, so the sign comes from $x^2 - 1$ alone.

$x$	$\cdots$	$-1$	$\cdots$	$1$	$\cdots$
$f'$	$+$	$0$	$-$	$0$	$+$
$f$	increasing	maximum $-2$	decreasing	minimum $2$	increasing

The decreasing stretch is interrupted at $x = 0$ , which lies outside the domain. The second derivative $f''(x) = \dfrac{2}{x^3}$ makes the curve concave up for $x > 0$ and concave down for $x < 0$ , with no inflection point.

Arithmetic and geometric means

For $x > 0$ the inequality between the two means gives the following¹.

x + \frac{1}{x} \geq 2\sqrt{x \cdot \frac{1}{x}} = 2

Equality holds only when $x = \dfrac{1}{x}$ , that is at $x = 1$ . The minimum value found by differentiation is precisely the case of equality in this inequality.

Range

Because the function is odd, the branch with $x < 0$ has a maximum of $-2$ , so no value in $-2 < y < 2$ is attained. The same conclusion follows algebraically: rearranging $x + \dfrac{1}{x} = k$ gives $x^2 - kx + 1 = 0$ , which has a real solution only when $k^2 - 4 \geq 0$ .

Asymptotes

Approach	Behaviour
$x \to 0^{+}$	$y \to +\infty$
$x \to 0^{-}$	$y \to -\infty$
$x \to \pm\infty$	$y - x \to 0$

The $y$ -axis is a vertical asymptote, and since $y - x = \dfrac{1}{x}$ tends to $0$ , the line $y = x$ is a slant asymptote that the curve hugs ever more closely far from the origin.

What the curve really is

Multiplying $y - x = \dfrac{1}{x}$ through by $x$ gives $x^2 - xy + 1 = 0$ . This is a conic, and the presence of the $xy$ term identifies it as a hyperbola. Its asymptotes are the $y$ -axis and the line $y = x$ , which meet at $45^\circ$ , so unlike the plain reciprocal it is not a rectangular hyperbola.

Invariance under reciprocals

Since $f\!\left(\dfrac{1}{x}\right) = f(x)$ , the function takes the same value at $x$ and at $\dfrac{1}{x}$ : for instance $f(2) = f\!\left(\dfrac{1}{2}\right) = \dfrac{5}{2}$ , so reciprocal pairs sit at the same height. Substituting $x = e^{t}$ gives $y = 2\cosh t$ , so the branch with $x > 0$ can be read as a stretched hyperbolic cosine.

Applications

A rectangle of area $1$ with height $x$ has width $\dfrac{1}{x}$ and perimeter $2\left(x + \dfrac{1}{x}\right) \geq 4$ . Equality occurs at $x = 1$ , showing that among rectangles of fixed area the square has the shortest perimeter.

Extended to complex numbers, $w = z + \dfrac{1}{z}$ is the Joukowsky transform², which maps a circle to an aerofoil section and is used in aerodynamics.

AM-GM inequality, Wikipedia
Joukowsky transform, Wikipedia

Graph of the Rational Function y=x+1xy = x + \dfrac{1}{x}y=x+x1​