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Power and Root Functions

MATH161
Reference: Stewart 1.2  •  Chapter: 1  •  Section: 2

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Power and Root Functions

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.

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

Prerequisites
Polynomial FunctionsExponent Rules
This skill
Power and Root Functions

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 (√x):          Odd root (∛x):
    |  /                      /
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    |/                  ────/────
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    0                  /

Reciprocal Functions (Negative Exponents)

Negative exponents produce reciprocal behavior:

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

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

        |
     \  |  /
      \ | /
───────\|/───────
       /|\
      / | \
        |

The Inverse Square Law

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

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 $x^a$:

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

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

  1. At a distance of 2 meters, a lamp provides 200 lux of illumination. Find $k$.
  2. What is the illumination at 4 meters?
  3. 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 $h(x) = x^{1/2}$.

  1. Find all points where any two of these functions intersect (besides the origin).
  2. For $0 < x < 1$, order the three functions from smallest to largest.
  3. For $x > 1$, order the three functions from smallest to largest.
  4. 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:

$$T^2 = a^3$$

  1. Express $T$ as a power function of $a$.
  2. Express $a$ as a power function of $T$.
  3. Earth has $a = 1$ AU and $T = 1$ year. Mars has $a = 1.52$ AU. Find Mars's orbital period.
  4. A comet has an orbital period of 76 years. Find its semi-major axis.
  5. 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.

Mastery Checklist

Mental Model

Think of the exponent as a "dial" that controls shape:

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


Connections

Looking back:

Looking ahead:

Real-world connections:


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Polynomial Functions Section Index Recognizing Function Families

Last updated: 2026-01-22