lecture 6: Root and Rational Functions — Graphs, Domain and Range, Discontinuities, and Infinity Behavior

Root Functions

1. What Is a Root Function?

A root function involves square roots, cube roots, or other fractional exponents:

$$f(x) = \sqrt{x}, \quad f(x) = \sqrt[3]{x}, \quad f(x) = x^{1/n}$$

• Square roots only work on non-negative inputs.
• Odd roots (like cube roots) work on all real numbers.

2. Domain and Range

Even roots need $x \geq 0$, odd roots are fine for any $x$.

3. Graphs of Root Functions

• Square Root: starts at (0,0), increases slowly
• Cube Root: passes through (0,0), symmetric across origin

Rational Functions

1. What Is a Rational Function?

A rational function is a ratio of two polynomials:

$$f(x) = \frac{P(x)}{Q(x)}$$

Examples:

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

2. Domain and Discontinuity

Rational functions are undefined where the denominator is zero.

Example:

$$f(x) = \frac{1}{x - 2} \quad \Rightarrow \quad \text{Domain: } x \neq 2$$

3. Types of Discontinuities

a) Removable (a "hole"):

Happens when a factor cancels from top and bottom.

$$f(x) = \frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad \text{but } x \neq 1$$

Hole at $x = 1$

b) Essential (vertical asymptote):

Occurs when the denominator goes to zero and does not cancel.

$$f(x) = \frac{1}{x - 3} \Rightarrow \text{Vertical asymptote at } x = 3$$

Asymptotes and Graph Behavior

1. Vertical Asymptotes

• Occur where the denominator = 0 and does not cancel.

2. Horizontal & Slant Asymptotes

3. Examples

Example 1:
Left side: $f(x)=\frac{x^3-2x^2+3x-1}{x-1}$
Domain: $x \neq 1$
Vertical asymptote: $x = 1$
Oblique asymptote: $y = x^2 - x + 2$

Right side: $f(x)=\frac{x^4-3x^2+2x}{x^2+1}$
Domain: $\mathbb{R}$
No vertical asymptotes
Oblique asymptote: $y = x^2 - 4$

Example 2:
$f(x)=\frac{2x^2+3x-1}{x^2-4}$
Domain: $x \neq \pm 2$
Vertical asymptotes: $x = \pm 2$
Horizontal asymptote: $y = 2$

Infinity Behavior

Root Functions

• As $x \to \infty$, root functions grow — but slowly.
  • $\sqrt{x} \to \infty$
  • $\sqrt[3]{x} \to \infty$ as $x \to \infty$, $\to -\infty$ as $x \to -\infty$

Rational Functions

• As $x \to \pm \infty$, rational functions approach asymptotes (if they exist).
  • e.g., $f(x) = \frac{3x^2}{x^2 + 1} \to 3$

Regular Exercises

1. Graph Analysis

Study the graph below and answer the questions about this rational function:

2. Types of Discontinuity

Determine whether the discontinuity is removable or essential:

3. Domain and Discontinuity

Find the domain of each function and describe any discontinuities:

• a) $f(x) = \sqrt{x - 3}$
• b) $f(x) = \frac{2x + 3}{x^2 - 4}$
• c) $f(x) = \frac{x^2 - 4}{x - 2}$

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

• A square root is only defined when the expression inside is ≥ 0.
• So:

$$x - 3 \geq 0 \Rightarrow x \geq 3$$

• Domain: $( [3, \infty) )$
• Discontinuity: None; the function is continuous for all values in its domain.

b) $( f(x) = \frac{2x + 3}{x^2 - 4} )$

• Denominator cannot be zero:

$$x^2 - 4 = 0 \Rightarrow x = \pm2$$

• Domain: $( \mathbb{R} \setminus \{-2, 2\} )$
• Discontinuity: Vertical asymptotes at $( x = -2 ) and ( x = 2 )$

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

• Factor numerator: $( (x - 2)(x + 2) )$
• The ( x - 2 ) cancels, leaving $( f(x) = x + 2 )$, but the original function is undefined at ( x = 2 ).
• Domain: $( \mathbb{R} \setminus \{2\} )$
• Discontinuity: Removable discontinuity (hole) at ( x = 2 )

4. Root Domain Puzzle

For each function, find the domain, but without graphing. Justify your reasoning algebraically:

• a) $f(x) = \sqrt{4 - x^2}$
• b) $g(x) = \sqrt{x^2 - 4}$
• c) $h(x) = \sqrt{(x - 1)(x + 2)}$

Solution
a) $( f(x) = \sqrt{4 - x^2} )$

• Inside square root must be ≥ 0:

$$4 - x^2 \geq 0 \Rightarrow -2 \leq x \leq 2$$

• Domain: ( [-2, 2] )

b) $( g(x) = \sqrt{x^2 - 4} )$

• Square root is defined when:

$$x^2 - 4 \geq 0 \Rightarrow x \leq -2 \text{ or } x \geq 2$$

• Domain: $( (-\infty, -2] \cup [2, \infty) )$

c) $( h(x) = \sqrt{(x - 1)(x + 2)} )$

• Find where the expression is ≥ 0. Use sign chart:
• Critical points: ( x = -2 ) and ( x = 1 )
• Test intervals:
  • ( x < -2 ): both factors negative → product positive
  • ( -2 < x < 1 ): one factor negative → product negative
  • ( x > 1 ): both positive → product positive
• Include points where the expression = 0
• Domain: $( (-\infty, -2] \cup [1, \infty) )$

5. Impossible Graph?

Here is a graph sketch:

• It has a hole at $x = 2$

• It has a vertical asymptote at $x = -1$

• It has a horizontal asymptote at $y = 3$

  • a) Can you write a rational function that fits this behavior?
  • c) What happens if you remove the factor that causes the hole?

Solution
a) We need a rational function with:

• Hole at ( x = 2 ): this means the factor ( (x - 2) ) is canceled
• Vertical asymptote at ( x = -1 ): the factor ( (x + 1) ) is in the denominator but not canceled
• Horizontal asymptote at ( y = 3 ): degree of numerator = degree of denominator; leading coefficients ratio = 3

One possible function:

$$f(x) = \frac{3(x - 2)}{(x - 2)(x + 1)} = \frac{3}{x + 1} \text{ for } x \neq 2$$

• Hole at ( x = 2 ), vertical asymptote at ( x = -1 ), and horizontal asymptote at ( y = 3 ) ✔️

b) If we remove the factor that causes the hole (i.e., cancel ( (x - 2) ) from numerator and denominator), the hole disappears.
The function becomes:

$$f(x) = \frac{3}{x + 1}$$

Now it is defined at $( x = 2 )$ and is no longer discontinuous there.

6. Graph from Clues

Build a rational function that:

• Has a vertical asymptote at $x = 2$

• Has a hole at $x = -1$

• Has a horizontal asymptote at $y = 1$

• Crosses the x-axis at $x = 3$

  • a) Write a possible expression.
  • b) Label all key features on a sketch.
  • c) Change your function so that the horizontal asymptote is now $y = 5$. What did you change?

Solution
a) Features needed:

• Vertical asymptote at ( x = 2 ): means factor ( (x - 2) ) in denominator
• Hole at ( x = -1 ): means factor ( (x + 1) ) in both numerator and denominator
• Horizontal asymptote at ( y = 1 ): numerator and denominator same degree; leading coefficients equal
• Zero at ( x = 3 ): factor ( (x - 3) ) in numerator

A possible function:

$$f(x) = \frac{(x - 3)(x + 1)}{(x - 2)(x + 1)}$$

• The ( (x + 1) ) cancels, creating a hole at ( x = -1 )

b) Key features on the graph:

• Hole at ( x = -1 )
• Vertical asymptote at ( x = 2 )
• x-intercept at ( x = 3 )
• Horizontal asymptote at ( y = 1 ) (degrees equal, leading coefficients = 1)

c) To change the horizontal asymptote to ( y = 5 ), make the leading coefficient of the numerator 5:

$$f(x) = \frac{5(x - 3)(x + 1)}{(x - 2)(x + 1)}$$

After canceling, this becomes:

$$f(x) = \frac{5(x - 3)}{x - 2}$$

Now the degrees are still equal, but the ratio of leading coefficients is 5, so the horizontal asymptote is ( y = 5 ).

Extra Advanced Exercises:

1. Asymptotic Limit Puzzle

Let

$$f(x) = \frac{\sqrt{x^2 + 5} - x}{x}$$

• a) Find the limit as $( x \to \infty )$ using rationalization
• b) Does this function have a horizontal asymptote?
• c) Analyze symmetry and behavior as $( x \to -\infty )$

2. Rational Function with Parameters

Let

$$f(x) = \frac{x^2 + ax + b}{x^2 - 4}$$

• a) Find conditions on ( a ) and ( b ) such that:
  • The function has no x-intercepts
  • A vertical asymptote at ( x = 2 )
  • A horizontal asymptote at ( y = 1 )
• b) Based on those conditions, determine the range of ( f(x) )

3. Domain and Discontinuity of a Root-Rational Combo

Let

$$f(x) = \frac{\sqrt{x^2 - 9}}{x^2 - 16}$$

• a) Determine the domain algebraically
• b) Identify and classify all discontinuities
• c) Analyze end behavior as $( x \to \pm\infty )$