The square root and reciprocal functions
What does a graph look like when its rule refuses some inputs, such as a negative under a square root or a zero in a denominator?
Where you are 3 in Parent functions
Why this exists
The parents met so far, the line y = x, the parabola y = x^2 and the cubic y = x^3 and the V of y = |x|, accept every real number. Their graphs run unbroken from far left to far right. Two of the most common operations do not behave that way. A square root undoes squaring, and no real number squares to a negative, so \displaystyle \sqrt{x} refuses every negative input. Division by zero has no answer, so \displaystyle \frac{1}{x} refuses the input 0.
Each refusal shows on the graph. The graph of \displaystyle y = \sqrt{x} begins at a point and goes only one way. The graph of \displaystyle y = \frac{1}{x} splits into two pieces that hug the axes without touching them. With these two, the list of Parent functions is long enough to set side by side, and that comparison is what lets you name a family from a graph, a table or a rule at a glance.
A sound engineer meets both: a tone’s period is the reciprocal of its frequency, and an amplitude factor is the square root of a power factor.
Definition
The square root function is \displaystyle f(x) = \sqrt{x}. For x \ge 0, \displaystyle \sqrt{x} is the one number that is at least 0 and squares to x, as in Square and cube roots. Its domain is x \ge 0.
The reciprocal function is \displaystyle g(x) = \frac{1}{x}. For x \ne 0, \displaystyle \frac{1}{x} is the one number that multiplied by x gives 1. Its domain is every real number except 0.
An asymptote of a graph is a line the graph comes as close to as you like, without end, as it runs off along it (what an asymptote is). The graph of \displaystyle \frac{1}{x} has two: the y-axis, which is the line x = 0, and the x-axis, which is the line y = 0.
Intuition
The square root graph is half a parabola on its side. On y = x^2 the points (1, 1), (2, 4) and (3, 9) say that 2 squares to 4 and 3 to 9. The square root reads each fact backwards, so its graph holds (1, 1), (4, 2) and (9, 3): the same points with the coordinates swapped. Swapping coordinates reflects a picture in the line y = x, so the graph of \displaystyle \sqrt{x} is the right half of the parabola reflected in that line. Only the right half, because \displaystyle \sqrt{x} is never negative.
It starts at the endpoint (0, 0) and always rises, but each new unit of height costs more width. Height 1 needs x to reach 1, height 2 needs 4, height 3 needs 9. The widths 1, 3, 5 keep growing, so the curve flattens as it goes, though it never stops rising.
The reciprocal graph trades big for small. A large input gives a small output and a small positive input gives a large output: the coordinates of every point multiply to 1, so as one shrinks toward 0 the other must grow. Near the y-axis the graph shoots up; far to the right it sinks toward the x-axis; it reaches neither. A negative input gives a negative output, so the left piece is the right piece turned half a turn about the origin.
Results
The square root function. Domain x \ge 0, range y \ge 0. The graph starts at (0, 0), passes through (1, 1), (4, 2) and (9, 3), and rises at every step, more and more slowly. For 0 < x < 1, \displaystyle \sqrt{x} > x; for x > 1, \displaystyle \sqrt{x} < x; they are equal only at x = 0 and x = 1.
The reciprocal function. Domain x \ne 0, range y \ne 0. The graph has two pieces, in the first and third quadrants, through (1, 1) and (-1, -1). Its asymptotes are x = 0 and y = 0. Each piece falls from left to right. The graph is its own mirror image in the line y = x, and its own half turn about the origin, a kind of symmetry studied later.
Side by side. The six parents so far:
| Parent | Domain | Range | Shape |
|---|---|---|---|
| y = x | all | all | line, slope 1 |
| y = x^2 | all | y \ge 0 | U, lowest at (0, 0) |
| y = x^3 | all | all | S, flat at (0, 0) |
| y = |x| | all | y \ge 0 | V, corner at (0, 0) |
| \displaystyle y = \sqrt{x} | x \ge 0 | y \ge 0 | half curve from (0, 0) |
| \displaystyle y = \frac{1}{x} | x \ne 0 | y \ne 0 | two pieces |
Every one passes through (1, 1), so that point never tells parents apart. The outputs at x = -1 and x = 2, taken together, are different for all six.
Why it is true
The square root always rises. Take 0 \le a < b. If \displaystyle \sqrt{a} \ge \sqrt{b}, squaring both sides (which keeps the direction, as both are at least 0) gives a \ge b, which is false. So \displaystyle \sqrt{a} < \sqrt{b}.
It rises more and more slowly. To climb from height n to height n + 1 the input must go from n^2 to (n + 1)^2, a width of (n + 1)^2 - n^2 = 2n + 1, which grows with n.
Where \displaystyle \sqrt{x} beats x. Write \displaystyle x = \sqrt{x} \cdot \sqrt{x}. If 0 < x < 1 then \displaystyle 0 < \sqrt{x} < 1, since \displaystyle \sqrt{1} = 1 and the square root rises, so \displaystyle x = \sqrt{x} \cdot \sqrt{x} < \sqrt{x} \cdot 1 = \sqrt{x}. If x > 1 then \displaystyle \sqrt{x} > 1 and the same step gives \displaystyle x > \sqrt{x}.
The reciprocal never gives 0, and gives everything else. If \displaystyle \frac{1}{x} = 0, multiplying by x gives 1 = 0, which is impossible. Any c \ne 0 is reached at \displaystyle x = \frac{1}{c}. So the range is y \ne 0.
The axes are asymptotes. For x > 0 and any number N > 0, \displaystyle \frac{1}{x} > N exactly when \displaystyle x < \frac{1}{N}, so inputs close enough to 0 lift the graph above any height you name, beside the y-axis. Likewise \displaystyle 0 < \frac{1}{x} < \frac{1}{N} exactly when x > N, so far enough right the graph is as close to the x-axis as you like, and never on it. Negative inputs do the same with signs reversed.
The symmetries. If (a, b) is on the graph then ab = 1. Then ba = 1, so (b, a) is on it, the mirror image in y = x; and (-a)(-b) = 1, so (-a, -b) is on it, the half turn.
Each piece falls. If 0 < a < b, dividing a < b by the positive number ab gives \displaystyle \frac{1}{b} < \frac{1}{a}. If a < b < 0, the product ab is again positive and the same step works.
Worked examples
Outputs at five inputs
Find \displaystyle \sqrt{x} and \displaystyle \frac{1}{x} at x = 9, 0.25, -4, \displaystyle \frac{1}{2} and 16.
SolutionThe square root asks which number at least 0 squares to the input. 3^2 = 9, so \displaystyle \sqrt{9} = 3. 0.5^2 = 0.25, so \displaystyle \sqrt{0.25} = 0.5. No real number squares to -4, since a square is never negative, so -4 is outside the domain. Next, \displaystyle \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.707. Last, 4^2 = 16, so \displaystyle \sqrt{16} = 4.
The reciprocal divides 1 by the input. \displaystyle \frac{1}{0.25} = 4, since \displaystyle 0.25 = \frac{1}{4} and \displaystyle 4 \cdot \frac{1}{4} = 1. Also \displaystyle \frac{1}{-4} = -0.25, \displaystyle \frac{1}{1/2} = 2 and \displaystyle \frac{1}{16} = 0.0625.
| x | 9 | 0.25 | -4 | \displaystyle \frac{1}{2} | 16 |
|---|---|---|---|---|---|
| \displaystyle \sqrt{x} | 3 | 0.5 | none | \displaystyle \frac{\sqrt{2}}{2} | 4 |
| \displaystyle \frac{1}{x} | \displaystyle \frac{1}{9} | 4 | -0.25 | 2 | \displaystyle \frac{1}{16} |
The two inputs between 0 and 1 come out larger under both rules. The reciprocal accepts -4, where the square root cannot.
Running each rule backwards
Find every x with \displaystyle \sqrt{x} = 7, with \displaystyle \sqrt{x} = 1.5 and with \displaystyle \sqrt{x} = -3. Then find every x with \displaystyle \frac{1}{x} = 5, with \displaystyle \frac{1}{x} = -0.2 and with \displaystyle \frac{1}{x} = 0.
SolutionIf \displaystyle \sqrt{x} = 7, squaring gives x = 49, and \displaystyle \sqrt{49} = 7 checks. In the same way \displaystyle \sqrt{x} = 1.5 gives x = 1.5^2 = 2.25. The equation \displaystyle \sqrt{x} = -3 has no solution, because the range is y \ge 0. Squaring would suggest x = 9, but \displaystyle \sqrt{9} = 3, not -3; the check is what catches it.
Multiplying \displaystyle \frac{1}{x} = 5 by x, allowed since x \ne 0, gives 1 = 5x, so x = 0.2. Likewise \displaystyle \frac{1}{x} = -0.2 gives 1 = -0.2x, so x = -5. And \displaystyle \frac{1}{x} = 0 has no solution, since 0 is not in the range.
On the graphs, a horizontal line y = c meets the square root graph once when c \ge 0 and the reciprocal graph once when c \ne 0; otherwise never.
How fast the square root climbs
How wide a stretch of inputs does \displaystyle \sqrt{x} need to rise from 3 to 4, and from 10 to 11? Then find \displaystyle \sqrt{x} at x = 0.01 and at x = 0.0001, and compare each output with its input.
SolutionHeight 3 is reached at x = 9 and height 4 at x = 16, a width of 16 - 9 = 7. Height 10 is reached at x = 100 and 11 at x = 121, a width of 21. Both match 2n + 1. Far out the graph is nearly flat.
Near 0 it is the other way. Since 0.1^2 = 0.01, \displaystyle \sqrt{0.01} = 0.1, ten times the input. Since 0.01^2 = 0.0001, \displaystyle \sqrt{0.0001} = 0.01, a hundred times the input. The segment from (0, 0) to (0.0001, 0.01) has slope \displaystyle \frac{0.01}{0.0001} = 100, so the curve leaves its endpoint almost straight up and then bends over.
The reciprocal near zero and far away
Make a table of \displaystyle \frac{1}{x} at x = 0.1, 0.01, 0.001 and at x = 10, 100, 1000, and describe the graph near the y-axis and far to the right. Then find every x > 0 with \displaystyle \frac{1}{x} > 500, and every x > 0 with \displaystyle \frac{1}{x} < 0.002.
Solution| x | 0.001 | 0.01 | 0.1 |
|---|---|---|---|
| \displaystyle \frac{1}{x} | 1000 | 100 | 10 |
| x | 10 | 100 | 1000 |
|---|---|---|---|
| \displaystyle \frac{1}{x} | 0.1 | 0.01 | 0.001 |
As x shrinks toward 0 through positive numbers, the outputs grow without bound: the graph climbs beside the y-axis, its vertical asymptote. As x grows, the outputs shrink toward 0 but stay positive: the graph sinks toward the x-axis, its horizontal asymptote.
For x > 0, multiplying \displaystyle \frac{1}{x} > 500 by x keeps the direction: 1 > 500x, so 0 < x < 0.002. In the same way \displaystyle \frac{1}{x} < 0.002 gives 1 < 0.002x, so x > 500.
Which parent made the table?
Each set of points comes from one of the six parents. Name it. (a) (0, 0), (0.25, 0.5), (4, 2), (16, 4). (b) (-4, -0.25), \displaystyle (-\frac{1}{2}, -2), \displaystyle (\frac{1}{2}, 2), \displaystyle (9, \frac{1}{9}). (c) (-1, -1), (0, 0), (1, 1), (2, 8). (d) (-1, 1), (0, 0), \displaystyle (\frac{1}{2}, \frac{1}{2}), (1, 1).
Solution(a) Each output squares to its input: 0.5^2 = 0.25, 2^2 = 4, 4^2 = 16. It is \displaystyle y = \sqrt{x}.
(b) Each pair multiplies to 1: (-4)(-0.25) = 1, \displaystyle (-\frac{1}{2})(-2) = 1, \displaystyle \frac{1}{2} \cdot 2 = 1, \displaystyle 9 \cdot \frac{1}{9} = 1. It is \displaystyle y = \frac{1}{x}.
(c) The first three points fit both y = x and y = x^3. The fourth decides: 2^3 = 8, while y = x would give 2. It is y = x^3.
(d) The points at -1, 0 and 1 fit both y = x^2 and y = |x|. At \displaystyle \frac{1}{2}, \displaystyle \left(\frac{1}{2}\right)^2 = \frac{1}{4} but \displaystyle \left|\frac{1}{2}\right| = \frac{1}{2}. It is y = |x|.
Period and frequency
Rania is a sound engineer. A tone of frequency f kilohertz makes 1000f vibrations a second, so one vibration takes \displaystyle \frac{1}{1000f} of a second, which is \displaystyle T = \frac{1}{f} milliseconds. Find the period of a 2.5 kHz tone and of a 16 kHz tone, and the frequency whose period is 0.8 ms. Her test oscillator runs from 0.02 kHz to 20 kHz. What periods can it produce?
Solution\displaystyle T = \frac{1}{2.5} = 0.4 ms, and \displaystyle T = \frac{1}{16} = 0.0625 ms. For a period of 0.8 ms, \displaystyle \frac{1}{f} = 0.8 gives \displaystyle f = \frac{1}{0.8} = 1.25 kHz; check: \displaystyle \frac{1}{1.25} = 0.8.
A frequency is positive, so only the right piece of the reciprocal graph is used: the situation cuts the domain to f > 0. That piece falls, so on 0.02 \le f \le 20 the longest period comes at the lowest frequency, \displaystyle \frac{1}{0.02} = 50 ms, and the shortest at the highest, \displaystyle \frac{1}{20} = 0.05 ms. The periods are 0.05 \le T \le 50. Doubling a frequency halves its period, since \displaystyle \frac{1}{2f} = \frac{1}{2} \cdot \frac{1}{f}.
Boundaries
\displaystyle \sqrt{x} means only the root that is at least 0. The equation x^2 = 9 has two solutions, 3 and -3, but \displaystyle \sqrt{9} is 3 alone; that choice is what makes \displaystyle \sqrt{x} a function.
The domain x \ge 0 is for real outputs. In the Complex numbers a negative number does have square roots, but these graphs plot real numbers only.
\displaystyle \frac{1}{x} falls on each piece separately, not across the gap at 0. A statement proved for x > 0 has to be checked again for x < 0.
The parent graph never reaches its asymptotes, but that is not part of what an asymptote is: other graphs can cross a horizontal asymptote, as Rational functions show.
Where this goes next
The next lessons move any parent: shifting, reflecting and stretching it, then combining the moves and drawing a graph from its equation. For these two parents, graphing square root functions follows the endpoint as it moves, and transforming the reciprocal graph follows both asymptotes. The square root opens the family of Radical functions. The reciprocal is the simplest of the Rational functions, and its stretched copies describe Inverse, joint and combined variation. Swapping coordinates is the idea behind Inverse functions.
Exercises
Outputs and inputs
- Find \displaystyle \sqrt{x} at x = 36, 0.09, 1.21 and 0.
- Find \displaystyle \frac{1}{x} at x = 1.6, -0.04, \displaystyle -\frac{1}{3} and \displaystyle \frac{2}{7}.
- Of the inputs -7, 0 and 0.3, which are in the domain of \displaystyle \sqrt{x}, and which in the domain of \displaystyle \frac{1}{x}?
- Find every x with \displaystyle \sqrt{x} = 12, with \displaystyle \sqrt{x} = 0.4 and with \displaystyle \sqrt{x} = -1.
- Find every x with \displaystyle \frac{1}{x} = 12.5 and with \displaystyle \frac{1}{x} = -6.
- Which of these points lie on the graph of \displaystyle y = \frac{1}{x}: (6.25, 0.16), (3, 0.3), \displaystyle (-0.75, -\frac{4}{3}), (1, -1)?
Shapes and comparisons
- How wide a stretch of inputs does \displaystyle \sqrt{x} need to rise from 5 to 6? From 12 to 13?
- Put \displaystyle \sqrt{x}, x and x^2 in order of size at x = 0.64 and at x = 2.89.
- Find every x > 0 with \displaystyle \frac{1}{x} > 200, and every x > 0 with \displaystyle \frac{1}{x} < 0.008.
- Find every x < 0 with \displaystyle \frac{1}{x} < -250.
- Name the parent behind each set of points: first \displaystyle (-3, -\frac{1}{3}), (-1, -1), (1, 1), \displaystyle (3, \frac{1}{3}); then (0, 0), (1, 1), (6.25, 2.5), (12.25, 3.5).
- Of the six parents x, x^2, x^3, |x|, \displaystyle \sqrt{x} and \displaystyle \frac{1}{x}, which have every real number as both domain and range? Which have range y \ge 0? Which graph has an endpoint?
- A graph lies wholly in the first and third quadrants and touches neither axis. Which parent is it? Which of the six parents have points in the second quadrant?
The sound engineer
- Nadir records a 3.2 kHz tone and a 0.0125 kHz hum. Find the period of each, using \displaystyle T = \frac{1}{f}.
- Which frequencies have periods of 0.32 ms and 40 ms?
- Erlan’s synthesizer plays from 0.016 kHz to 8 kHz. What periods can it produce?
- When a signal’s power changes by a factor P, its amplitude changes by the factor \displaystyle A = \sqrt{P}. Rasa raises the power of one track by a factor of 25 and of another by 1.44. By what factor does each amplitude change?
- By what factor must the power rise to make the amplitude 8 times as large?
- Nkem cuts a track’s power to 0.81 of what it was. By what factor does the amplitude change, and why does it fall by less than the power?
Reasoning
- Explain why the graph of \displaystyle y = \frac{1}{x} has no point on either axis.
- Show that the graph of \displaystyle y = \sqrt{x} is exactly the part of y = x^2 with x \ge 0, with each point’s coordinates swapped.
- Erlan says \displaystyle \frac{1}{x} always gets smaller as x gets larger. Test the claim with x = -2 and x = 3, and say what is true instead.
Answers
- 6, 0.3, 1.1 and 0, since 6^2 = 36, 0.3^2 = 0.09, 1.1^2 = 1.21, 0^2 = 0, and each root is at least 0.
- 0.625, -25, -3 and 3.5. Dividing 1 by \displaystyle -\frac{1}{3} gives -3, and by \displaystyle \frac{2}{7} gives \displaystyle \frac{7}{2}.
- \displaystyle \sqrt{x}: 0 and 0.3, not -7. \displaystyle \frac{1}{x}: -7 and 0.3, not 0.
- x = 144 and x = 0.16. \displaystyle \sqrt{x} = -1 has no solution, since \displaystyle \sqrt{x} \ge 0.
- x = 0.08 and \displaystyle x = -\frac{1}{6}, from 1 = 12.5x and 1 = -6x.
- (6.25, 0.16) and \displaystyle (-0.75, -\frac{4}{3}) only: their coordinates multiply to 1. But 3 \cdot 0.3 = 0.9 and 1 \cdot (-1) = -1.
- 11 and 25. Height 5 to 6 needs x from 25 to 36; height 12 to 13 needs 144 to 169. Both match 2n + 1.
- At 0.64: \displaystyle x^2 = 0.4096 < x < \sqrt{x} = 0.8. At 2.89: \displaystyle \sqrt{x} = 1.7 < x < x^2 = 8.3521. The order flips at x = 1.
- 0 < x < 0.005 and x > 125. Multiplying by the positive x gives 1 > 200x and 1 < 0.008x.
- -0.004 < x < 0. Multiplying by the negative x reverses the inequality: 1 > -250x, so x > -0.004.
- \displaystyle y = \frac{1}{x}, since each pair multiplies to 1; then \displaystyle y = \sqrt{x}, since 2.5^2 = 6.25 and 3.5^2 = 12.25.
- Domain and range all reals: x and x^3. Range y \ge 0: x^2, |x| and \displaystyle \sqrt{x}. An endpoint: only \displaystyle \sqrt{x}, at (0, 0).
- \displaystyle y = \frac{1}{x}; the cubic also lies in those quadrants but passes through the origin. The second quadrant holds points of x^2 and |x|, where a negative input gives a positive output.
- 0.3125 ms and 80 ms: \displaystyle \frac{1}{3.2} = 0.3125 and \displaystyle \frac{1}{0.0125} = 80.
- 3.125 kHz and 0.025 kHz, since \displaystyle f = \frac{1}{T}: \displaystyle \frac{1}{0.32} = 3.125 and \displaystyle \frac{1}{40} = 0.025.
- 0.125 \le T \le 62.5 ms. The piece falls, so the highest frequency gives the shortest period, \displaystyle \frac{1}{8} = 0.125, and the lowest the longest, \displaystyle \frac{1}{0.016} = 62.5.
- By 5 and by 1.2: \displaystyle \sqrt{25} = 5 and \displaystyle \sqrt{1.44} = 1.2.
- By 64, since 8^2 = 64 and so \displaystyle \sqrt{64} = 8.
- By 0.9, since 0.9^2 = 0.81. For 0 < x < 1, \displaystyle \sqrt{x} > x, so the factor for the amplitude is closer to 1 than the factor for the power.
- A point on the y-axis has x = 0, which is outside the domain. A point on the x-axis has y = 0, and \displaystyle \frac{1}{x} = 0 would mean 1 = 0 \cdot x = 0.
- A point of y = x^2 with x \ge 0 is (a, a^2) with a \ge 0; swapped, it is (a^2, a). Since a is at least 0 and squares to a^2, \displaystyle \sqrt{a^2} = a, so (a^2, a) is on \displaystyle y = \sqrt{x}. Each point \displaystyle (x, \sqrt{x}) arises this way, from \displaystyle a = \sqrt{x}.
- -2 < 3, yet \displaystyle \frac{1}{-2} = -0.5 is less than \displaystyle \frac{1}{3}: the output went up. What is true: \displaystyle \frac{1}{x} falls on x < 0 and falls on x > 0, but the outputs jump from negative to positive across the gap at 0.