The Reciprocal-Square Function, foundations for calculus lesson

Stand twice as far from a lamp and the light that reaches you drops to a quarter, not a half. The pull of gravity and the intensity of sound from a small source behave the same way. All of them follow one function, one over the square of the distance, and its graph is simple once you build it in the right order. First square the input, then take the reciprocal.

Key terms

Reciprocal-square function

Definition The reciprocal-square function is the function $f$ given by

f(x) = \frac{1}{x^2}, \qquad x \ne 0.

The expression $\frac{1}{x^2}$ means $\frac{1}{(x^2)}$ . To evaluate it you square $x$ first and then take the reciprocal of that square. Because the reciprocal of a square equals the square of the reciprocal when $x \ne 0$ , the same value also equals $\left(\frac{1}{x}\right)^2$ .

In plain words Square the number, then flip it into one over that square. Squaring wipes out the sign, and flipping turns large squares into small outputs and small squares into large ones.

Example Here are a few values, built in two steps.

$x$	$x^2$	$f(x) = 1/x^2$
$-2$	$4$	$1/4$
$-1$	$1$	$1$
$-1/2$	$1/4$	$4$
$1/2$	$1/4$	$4$
$1$	$1$	$1$
$2$	$4$	$1/4$

Plotting these points already shows two curves, one on each side of the $y$ -axis, both above the $x$ -axis.

Negative integer power

Definition For a real number $x \ne 0$ and a positive integer $n$ ,

x^{-n} = \frac{1}{x^n}.

The expression $0^{-n}$ is undefined, since it would require dividing by $0$ . In particular the reciprocal-square function is the power function $f(x) = x^{-2}$ with domain all $x \ne 0$ . The minus sign in the exponent means reciprocal. It does not make the output negative.

In plain words A negative exponent tells you to put the power in the denominator. So $x^{-2}$ is one over $x^2$ , which is a positive number for every nonzero $x$ .

Example The value at $x = -3$ is

(-3)^{-2} = \frac{1}{(-3)^2} = \frac{1}{9},

which is positive. Compare this with $-3^{-2}$ , which means $-(3^{-2}) = -\frac{1}{9}$ because the exponent applies before the minus sign. Looking ahead, calculus often prefers the form $x^{-2}$ , because its rules for powers then apply to this function without any rewriting.

Positive outputs

Definition For every $x \ne 0$ the square $x^2$ is strictly positive, and the reciprocal of a positive number is positive, so

\frac{1}{x^2} > 0 \qquad \text{for every } x \ne 0.

No input gives the output $0$ or a negative output. The graph lies entirely above the $x$ -axis and never touches it.

In plain words Squaring can never give a negative number, and with $0$ excluded it cannot give zero either. One over a positive number stays positive.

Example The equation $\frac{1}{x^2} = -4$ has no real solution, because the left side is positive for every allowed $x$ . You can see this before doing any algebra. Multiplying both sides by $x^2$ would give $1 = -4x^2$ , which also fails for every real $x$ , confirming the answer.

Domain and range

Definition The domain of $f(x) = \frac{1}{x^2}$ is the set of all real $x$ with $x \ne 0$ , which is $(-\infty, 0) \cup (0, \infty)$ . The range is $(0, \infty)$ . Every positive number $y$ is an output, and it is reached at exactly two inputs,

x = \frac{1}{\sqrt{y}} \quad \text{and} \quad x = -\frac{1}{\sqrt{y}}.

In plain words You can put in anything except $0$ , and you get out every positive number and nothing else. Each positive output appears twice, once on each side.

Example To find where the output equals $9$ , solve $\frac{1}{x^2} = 9$ . That gives $x^2 = \frac{1}{9}$ , so

x = \frac{1}{3} \quad \text{or} \quad x = -\frac{1}{3}.

Both check in the original equation, since $(\pm \frac{1}{3})^2 = \frac{1}{9}$ and its reciprocal is $9$ . This matches the formula with $y = 9$ and $\sqrt{9} = 3$ .

Even symmetry

Definition A function $g$ is even when its domain is symmetric about $0$ and $g(-x) = g(x)$ for every $x$ in the domain. The reciprocal-square function is even, because for every $x \ne 0$

f(-x) = \frac{1}{(-x)^2} = \frac{1}{x^2} = f(x).

The graph of an even function is symmetric across the $y$ -axis.

In plain words Inputs $x$ and $-x$ have the same square, so they get the same output. The left half of the graph is the mirror image of the right half.

Example The inputs $-4$ and $4$ both square to $16$ , so both outputs are $\frac{1}{16}$ . Contrast this with the reciprocal function $\frac{1}{x}$ , which gives $-\frac{1}{4}$ at $-4$ and $\frac{1}{4}$ at $4$ . That function is odd, its left branch sits below the axis, and its graph has symmetry through the origin rather than across the $y$ -axis. See the reciprocal function for that graph.

Two branches

Definition The graph of $f(x) = \frac{1}{x^2}$ consists of two separate pieces, called branches, one over the interval $(-\infty, 0)$ and one over $(0, \infty)$ . The function is increasing on $(-\infty, 0)$ and decreasing on $(0, \infty)$ . It is neither increasing nor decreasing on its whole domain. A context can restrict the domain, for example to $x > 0$ when $x$ is a distance, and then only the right branch belongs to the graph.

In plain words Coming in from the far left, the graph climbs as you approach $0$ . Leaving $0$ to the right, it falls. The two pieces never connect, because $0$ is not in the domain.

Example On the left branch, $-2 < -1$ and $f(-2) = \frac{1}{4} < 1 = f(-1)$ , so the outputs rise. On the right branch, $1 < 2$ and $f(1) = 1 > \frac{1}{4} = f(2)$ , so the outputs fall. Across the gap the pattern breaks. The inputs $-2 < 1$ give $f(-2) = \frac{1}{4} < f(1) = 1$ , an increase, while $1 < 2$ gave a decrease. That is why you describe the behavior one interval at a time.

Vertical asymptote

Definition The line $x = a$ is a vertical asymptote of the graph of $g$ when, as $x$ approaches $a$ from at least one side, the values $g(x)$ eventually exceed every positive number or eventually fall below every negative number. In symbols, $g(x) \to \infty$ or $g(x) \to -\infty$ from that side. For the reciprocal-square function the line $x = 0$ is a vertical asymptote, and the outputs grow without bound toward $+\infty$ from both sides,

\begin{gathered} \frac{1}{x^2} \to \infty \\[0.4em] \text{as } x \to 0 \text{ from either side}. \end{gathered}

In plain words Near $0$ the square is tiny and positive, so one over it is huge and positive. Both branches shoot upward along the $y$ -axis.

Example The output at $x = 0.01$ is $\frac{1}{0.0001} = 10000$ , and the output at $x = -0.01$ is the same. At $x = 0.001$ it is $1000000$ . Compare $\frac{1}{x}$ , which goes up on the right of $0$ and down on the left. For $\frac{1}{x^2}$ both sides go up, and this sign pattern is what you check when you study infinite limits.

Horizontal asymptote

Definition The line $y = b$ is a horizontal asymptote of the graph of $g$ when $g(x)$ gets as close to $b$ as you like for all sufficiently large $x$ , or for all sufficiently negative $x$ . For the reciprocal-square function the line $y = 0$ is a horizontal asymptote in both directions,

\begin{gathered} \frac{1}{x^2} \to 0 \\[0.4em] \text{as } x \to \infty \text{ and as } x \to -\infty. \end{gathered}

Since every output is positive, the graph approaches the axis from above and never reaches it.

In plain words Far from the origin the square is enormous, so one over it is tiny. The branches flatten out toward the $x$ -axis but always stay just above it.

Example The outputs at $x = 10$ and $x = 100$ are

\begin{gathered} f(10) = \frac{1}{100} = 0.01, \\[0.4em] f(100) = \frac{1}{10000} = 0.0001. \end{gathered}

The same values occur at $-10$ and $-100$ . The outputs shrink much faster than those of $\frac{1}{x}$ , which gives $0.01$ only at $x = 100$ .

Inverse-square change

Definition A quantity follows an inverse-square law when it has the form $g(x) = \frac{c}{x^2}$ for a constant $c \ne 0$ . For every $x \ne 0$ and every constant $k \ne 0$ ,

g(kx) = \frac{c}{k^2 x^2} = \frac{g(x)}{k^2}.

So multiplying the input by $k$ divides the output by $k^2$ . In particular, doubling the input divides the output by $4$ , and tripling it divides the output by $9$ .

In plain words The output does not just shrink in proportion to the input. It shrinks by the square of the factor, so changes in the input have a strong effect.

Example Suppose the light intensity at distance $d$ meters from a small lamp is

I(d) = \frac{360}{d^2}, \qquad d > 0.

At $d = 2$ the intensity is $90$ . Doubling the distance to $4$ gives $\frac{360}{16} = 22.5$ , which is $\frac{90}{4}$ . Tripling the distance from $2$ to $6$ gives $\frac{360}{36} = 10$ , which is $\frac{90}{9}$ . Only positive distances make sense here, so the left branch is not part of this model.

Check yourself

A quantity is modeled by $g(x) = \frac{50}{x^2}$ for $x > 0$ . Find $g(5)$ and $g(10)$ , and say how the output changes when the input doubles.

Show the answer

The values are $g(5) = \frac{50}{25} = 2$ and $g(10) = \frac{50}{100} = \frac{1}{2}$ . Doubling the input divides the output by $2^2 = 4$ , and indeed $\frac{1}{2} = \frac{2}{4}$ .

Find every real $x$ with $\frac{1}{x^2} = \frac{1}{4}$ , and every real $x$ with $\frac{1}{x^2} = 0$ .

Show the answer

The first equation gives $x^2 = 4$ , so $x = 2$ or $x = -2$ , and both check. The second has no solution, because $\frac{1}{x^2}$ is strictly positive for every $x \ne 0$ . The value $0$ is not in the range.

Is $f(x) = \frac{1}{x^2}$ decreasing on its whole domain? Explain with specific inputs.

Show the answer

No. It is decreasing on $(0, \infty)$ but increasing on $(-\infty, 0)$ . For instance $-2 < 1$ while $f(-2) = \frac{1}{4} < 1 = f(1)$ , so the outputs went up as the input went up.

The Reciprocal-Square Function

Key terms

Reciprocal-square function

Negative integer power

Positive outputs

Domain and range

Even symmetry

Two branches

Vertical asymptote

Horizontal asymptote

Inverse-square change

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus