Power and Root Functions

6 min read

Reference: Stewart §1.2

Beyond Integer Exponents

What happens when the exponent isn’t a positive integer? Polynomials use $x^2$, $x^3$, and so on, but nature often requires more flexibility.

Square roots appear when you solve $y^2 = x$ for $y$
Cube roots arise in volume problems where you know the volume but need the side length
Negative exponents describe inverse relationships like electrical resistance or gravitational force

All of these are power functions: $f(x) = x^a$ where $a$ can be any real number. Understanding how the exponent $a$ affects the shape, domain, and behavior of these functions is essential for modeling real phenomena.

Prerequisite Map

Quick Reference

Property	Value
Concept	Essential Functions
Chapter	Chapter 1, Section 2
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Power Function

A power function has the form:

$$f(x) = x^a$$

where $a$ is a constant (the exponent). The behavior depends entirely on the value of $a$.

Types of Power Functions

Exponent Type	Example	Name	Domain
Positive integer	$x^3$	Polynomial term	All real $x$
Positive fraction	$x^{1/2} = \sqrt{x}$	Root function	$x \geq 0$ (for even root)
Negative integer	$x^{-1} = \frac{1}{x}$	Reciprocal	$x \neq 0$
Negative fraction	$x^{-1/2} = \frac{1}{\sqrt{x}}$	Reciprocal root	$x > 0$

Root Functions

Root functions are power functions with fractional exponents:

$$\sqrt[n]{x} = x^{1/n}$$

Domain considerations:

Even root (like $\sqrt{x}$ or $\sqrt[4]{x}$): Domain is $x \geq 0$. You can’t take the square root of a negative number (in real numbers).
Odd root (like $\sqrt[3]{x}$ or $\sqrt[5]{x}$): Domain is all real numbers. Cube root of $-8$ is $-2$.

Even root (√x):          Odd root (∛x):
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    0                  /

Reciprocal Functions (Negative Exponents)

Negative exponents produce reciprocal behavior:

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

Key example: $f(x) = x^{-1} = \frac{1}{x}$

Domain: $x \neq 0$
As $x \to 0$, $f(x) \to \pm\infty$ (vertical asymptote)
As $x \to \pm\infty$, $f(x) \to 0$ (horizontal asymptote)

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───────\|/───────
       /|\
      / | \
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The Inverse Square Law

Many physical phenomena follow $f(x) = x^{-2} = \frac{1}{x^2}$:

Light intensity decreases as $1/r^2$ from the source
Gravitational force follows $F \propto 1/r^2$
Sound intensity decreases as $1/r^2$

Why? Energy spreads over a sphere’s surface. Surface area is $4\pi r^2$, so intensity per unit area decreases proportionally to $1/r^2$.

Comparing Power Functions Near Zero and at Infinity

For $x > 0$:

When $x$ is small $(0 < x < 1)$	When $x$ is large $(x > 1)$
Higher positive exponents give smaller values	Higher positive exponents give larger values
$x^3 < x^2 < x$ for $0 < x < 1$	$x^3 > x^2 > x$ for $x > 1$
Negative exponents give larger values	Negative exponents give smaller values

Practice Problems

Level 1 Rewriting with Exponents

Write each expression in the form $x^a$:

$\sqrt[3]{x}$
$\frac{1}{x^4}$
$\sqrt[5]{x^2}$
$\frac{1}{\sqrt{x}}$

Thought Process

Use these rules:

$\sqrt[n]{x} = x^{1/n}$
$\frac{1}{x^n} = x^{-n}$
$\sqrt[n]{x^m} = x^{m/n}$

Combine rules as needed.

Show Answer

(a) $\sqrt[3]{x} = \boxed{x^{1/3}}$

(b) $\frac{1}{x^4} = \boxed{x^{-4}}$

(c) $\sqrt[5]{x^2} = \boxed{x^{2/5}}$

(d) $\frac{1}{\sqrt{x}} = \frac{1}{x^{1/2}} = \boxed{x^{-1/2}}$

Level 2 Finding Domains

Find the domain of each function:

$f(x) = \sqrt{x - 3}$
$g(x) = \sqrt[3]{x - 3}$
$h(x) = \frac{1}{x^2 - 4}$
$k(x) = x^{-2/3}$

Thought Process

For each function, ask:

Does it have an even root? If so, the radicand must be $\geq 0$.
Does it have a denominator? If so, the denominator can’t be zero.
Is there a negative exponent? This creates a denominator.

Combine restrictions to find the domain.

Show Answer

(a) $f(x) = \sqrt{x - 3}$

Even root (square root), so we need $x - 3 \geq 0$, which gives $x \geq 3$.

Domain: $\boxed{[3, \infty)}$

(b) $g(x) = \sqrt[3]{x - 3}$

Odd root (cube root): no restrictions. Any real number has a cube root.

Domain: $\boxed{(-\infty, \infty)}$ or all real numbers

(c) $h(x) = \frac{1}{x^2 - 4}$

Denominator can’t be zero: $x^2 - 4 \neq 0$, so $x \neq \pm 2$.

Domain: $\boxed{(-\infty, -2) \cup (-2, 2) \cup (2, \infty)}$

(d) $k(x) = x^{-2/3} = \frac{1}{x^{2/3}} = \frac{1}{\sqrt[3]{x^2}}$

The cube root is defined for all $x$, but we can’t divide by zero. $\sqrt[3]{x^2} = 0$ only when $x = 0$.

Domain: $\boxed{(-\infty, 0) \cup (0, \infty)}$ or $x \neq 0$

Level 3 Inverse Square Law Application

The illumination $I$ from a light source varies inversely as the square of the distance $d$ from the source: $$I = \frac{k}{d^2}$$ where $k$ is a constant depending on the light’s brightness.

At a distance of 2 meters, a lamp provides 200 lux of illumination. Find $k$.
What is the illumination at 4 meters?
At what distance is the illumination 50 lux?

Thought Process

This is the inverse square law in action.

For part (a): Substitute the known values $(d, I) = (2, 200)$ and solve for $k$.

For parts (b) and (c): Use the formula with the known $k$ and solve for the unknown.

Show Answer

(a) Using $(d, I) = (2, 200)$: $$200 = \frac{k}{2^2} = \frac{k}{4}$$ $$k = 200 \times 4 = \boxed{800}$$

(b) At $d = 4$ meters: $$I = \frac{800}{4^2} = \frac{800}{16} = \boxed{50 \text{ lux}}$$

(c) Find $d$ when $I = 50$: $$50 = \frac{800}{d^2}$$ $$d^2 = \frac{800}{50} = 16$$ $$d = \boxed{4 \text{ meters}}$$

Observation: Doubling the distance (from 2m to 4m) reduces illumination by a factor of 4 (from 200 to 50 lux). This is characteristic of inverse square relationships.

Level 4 Comparing Growth Rates

Consider the functions $f(x) = x^2$, $g(x) = x^3$, and $h(x) = x^{1/2}$.

Find all points where any two of these functions intersect (besides the origin).
For $0 < x < 1$, order the three functions from smallest to largest.
For $x > 1$, order the three functions from smallest to largest.
Explain the pattern in terms of exponents.

Thought Process

To find intersections, set pairs of functions equal and solve. For example, $x^2 = x^3$ gives $x^2(1-x) = 0$, so $x = 0$ or $x = 1$.

For the ordering questions, test a specific value like $x = 0.5$ or $x = 2$.

The pattern relates to whether $x$ is less than or greater than 1, and how exponents amplify or compress values.

Show Answer

(a) Find intersections (excluding origin):

$f = g$: $x^2 = x^3 \Rightarrow x^2(1-x) = 0 \Rightarrow x = 1$ Point: $(1, 1)$

$f = h$: $x^2 = x^{1/2} \Rightarrow x^4 = x \Rightarrow x(x^3 - 1) = 0 \Rightarrow x = 1$ Point: $(1, 1)$

$g = h$: $x^3 = x^{1/2} \Rightarrow x^6 = x \Rightarrow x(x^5 - 1) = 0 \Rightarrow x = 1$ Point: $(1, 1)$

All three functions meet at $(1, 1)$ (besides the origin).

(b) For $0 < x < 1$, test $x = 0.5$:

$f(0.5) = 0.25$
$g(0.5) = 0.125$
$h(0.5) = \sqrt{0.5} \approx 0.707$

Order: $\boxed{g(x) < f(x) < h(x)}$ or $x^3 < x^2 < x^{1/2}$

(c) For $x > 1$, test $x = 2$:

$f(2) = 4$
$g(2) = 8$
$h(2) = \sqrt{2} \approx 1.414$

Order: $\boxed{h(x) < f(x) < g(x)}$ or $x^{1/2} < x^2 < x^3$

(d) The pattern:

For $0 < x < 1$: Larger exponents give smaller values. Multiplying a number less than 1 by itself makes it smaller.

For $x > 1$: Larger exponents give larger values. Multiplying a number greater than 1 by itself makes it larger.

The crossover happens at $x = 1$, where $1^a = 1$ for any $a$.

Level 5 Kepler's Third Law

Kepler’s Third Law states that for planets orbiting the Sun, the orbital period $T$ (in years) relates to the semi-major axis $a$ (average distance from Sun, in AU) by:

$$T^2 = a^3$$

Express $T$ as a power function of $a$.
Express $a$ as a power function of $T$.
Earth has $a = 1$ AU and $T = 1$ year. Mars has $a = 1.52$ AU. Find Mars's orbital period.
A comet has an orbital period of 76 years. Find its semi-major axis.
If a planet's distance from the Sun doubles, by what factor does its orbital period change?

Thought Process

From $T^2 = a^3$:

To solve for $T$: take the square root: $T = a^{3/2}$
To solve for $a$: take the cube root of both sides, then... $a = T^{2/3}$

For parts (c) and (d), substitute values into the appropriate formula.

For part (e), if $a$ doubles to $2a$, find $T(2a)/T(a)$.

Show Answer

(a) From $T^2 = a^3$: $$T = (a^3)^{1/2} = \boxed{a^{3/2}}$$

(b) From $T^2 = a^3$: $$a^3 = T^2$$ $$a = (T^2)^{1/3} = \boxed{T^{2/3}}$$

(c) For Mars with $a = 1.52$ AU: $$T = (1.52)^{3/2} = (1.52)^{1.5} = \sqrt{1.52^3} = \sqrt{3.51} \approx \boxed{1.87 \text{ years}}$$

(d) For the comet with $T = 76$ years: $$a = 76^{2/3} = (76^2)^{1/3} = \sqrt[3]{5776} \approx \boxed{17.9 \text{ AU}}$$

(e) If $a$ is replaced by $2a$: $$T_{new} = (2a)^{3/2} = 2^{3/2} \cdot a^{3/2} = 2\sqrt{2} \cdot T$$

The period increases by a factor of $\boxed{2\sqrt{2} \approx 2.83}$.

Physical insight: Farther planets move slower AND have longer paths. Both effects compound to make the period increase faster than just doubling.

Common Misconceptions

Common misconception

$\sqrt{x + y} = \sqrt{x} + \sqrt{y}$. The square root does not distribute over addition. A counterexample: $\sqrt{9 + 16} = \sqrt{25} = 5$, but $\sqrt{9} + \sqrt{16} = 3 + 4 = 7 \neq 5$. The root is a nonlinear operation, and breaking it across a sum produces a larger value than the correct answer. This error pattern (applying a nonlinear operation term-by-term) is one of the most common algebra mistakes in calculus.

Common misconception

$x^{-2}$ means $x$ squared with a negative value. A negative exponent denotes a reciprocal: $x^{-2} = 1/x^2$. It says nothing about the sign of $x$ or of the output. For $x = 3$: $3^{-2} = 1/9$, a positive number. For $x = -2$: $(-2)^{-2} = 1/4$, also positive. The negative in the exponent shifts from numerator to denominator, not from positive to negative output.

Mastery Checklist

I can rewrite root and reciprocal expressions using fractional/negative exponents
I can determine domains of power functions (even roots need non-negative input)
I can apply inverse square laws to physical problems
I understand how exponent size affects function values for $x < 1$ vs $x > 1$
I can work with power functions in modeling contexts (like Kepler’s law)

Mental Model

Think of the exponent as a “dial” that controls shape:

Dial at 1: Straight line through origin
Turn dial up (>1): Curve bends upward, steeper for large $x$
Turn dial down (0 < a < 1): Curve flattens, less steep for large $x$
Dial at 0: Horizontal line at $y = 1$
Dial negative: Flips to become $1/x^{\vert a\vert }$, shooting up near zero, flattening toward infinity

The sign of the exponent determines whether the function grows or decays as $x$ increases.

Connections

Looking back:

Polynomial functions are sums of positive-integer power functions
Exponent rules from algebra are essential for simplification

Looking ahead:

The power rule for derivatives says $\frac{d}{dx}x^a = ax^{a-1}$: the exponent comes down and decreases by 1
Rational functions are quotients of polynomials

Real-world connections:

Inverse square laws govern light, gravity, electrostatic force, sound
Kepler’s laws describe planetary motion
Allometric scaling in biology: metabolic rate scales as $(\text{mass})^{3/4}$

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Polynomial Functions	Section Index	Recognizing Function Families

Last updated: 2026-01-22