S4 Mathematics Practice Book · Unit 6: Polynomial, rational and irrational functions

Unit 6 Polynomial, rational and irrational functions

Key unit competence: Analyse polynomial, rational and irrational functions by determining their domains, representing them graphically and solving related problems.

Key ideas

Domain of definition: polynomials: $R$ ; rational $\frac{P ( x )}{Q ( x )}$ : $Q (x) \neq = 0$ ; irrational $n f (x)$ with $n$ even: $f (x) \geq 0$ (odd $n$ : domain of $f$ ).
Range: the set of all output values; found from the graph or by solving $y = f (x)$ for $x$ .
Operations: $(f \pm g) (x)$ , $(f g) (x)$ , $(\frac{f}{g}) (x)$ with $g (x) \neq = 0$ . Composite $(g \circ f) (x) = g (f (x))$ , defined where $f (x)$ is in the domain of $g$ .
Inverse $f^{- 1}$ exists for one-to-one (bijective) functions: swap $x$ and $y$ in $y = f (x)$ and solve for $y$ . The graphs of $f$ and $f^{- 1}$ are symmetric about $y = x$ .
Parity: $f$ is even if $f (- x) = f (x)$ (symmetric about the y-axis); odd if $f (- x) = - f (x)$ (symmetric about the origin). The domain must be symmetric.
Partial fractions: $\frac{p x + q}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}$ ; repeated factor: $\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}}$ ; quadratic factor: $\frac{B x + C}{x ^{2} + k}$ . If the degree of the numerator is not less, divide first.

Formulas to know

D_{f} = {x \in R : Q (x) \neq = 0} for \frac{P}{Q}

(g \circ f) (x) = g (f (x))

f (- x) = f (x) (even), f (- x) = - f (x) (odd)

\frac{p x + q}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}

Common misconceptions

"The domain of $x - 3$ is $x > 3$ ." No: $x \geq 3$ ( $0 = 0$ is fine).
"Every function is either even or odd." No: $f (x) = x + 1$ is neither.
" $f^{- 1} (x) = \frac{1}{f ( x )}$ ." No: the inverse function undoes $f$ ; $\frac{1}{f ( x )}$ is the reciprocal.

Worked examples

Example 1. Find the domain of $f (x) = \frac{x + 1}{x - 2}$ .

Solution

Need $x + 1 \geq 0$ and $x \neq = 2$ : $D_{f} = [- 1, 2 [\cup] 2, + \infty [$ .

Example 2. Decompose $\frac{5 x + 1}{( x - 1 ) ( x + 2 )}$ .

Solution

$5 x + 1 = A (x + 2) + B (x - 1)$ . $x = 1$ : $6 = 3 A$ , $A = 2$ . $x = - 2$ : $- 9 = - 3 B$ , $B = 3$ . So $\frac{2}{x - 1} + \frac{3}{x + 2}$ .

Practice questions

Easier and harder questions are mixed, just like in a real exam.

Find the domain of $f (x) = \frac{x + 1}{x - 2}$ .
Show answer
$x \geq - 1$ and $x \neq = 2$ : $[- 1, 2 [\cup] 2, + \infty [$ .
Common mistake: Forgetting to exclude 2.
Decompose $\frac{3 x + 5}{( x + 1 ) ^{2}}$ .
Show answer
$3 x + 5 = A (x + 1) + B$ : $A = 3$ , $B = 2$ . $\frac{3}{x + 1} + \frac{2}{( x + 1 ) ^{2}}$ .
Common mistake: Using only $\frac{A}{( x + 1 ) ^{2}}$ .
Given $f (x) = x^{2} - 1$ and $g (x) = \frac{1}{x}$ , find $(f \circ g) (x)$ and $(g \circ f) (x)$ with their domains.
Show answer
$f (g (x)) = \frac{1}{x ^{2}} - 1$ , $x \neq = 0$ . $g (f (x)) = \frac{1}{x ^{2} - 1}$ , $x \neq = \pm 1$ .
Common mistake: Assuming both have the same domain.
Find the range of $f (x) = x^{2} - 4 x + 7$ .
Show answer
$f (x) = (x - 2)^{2} + 3 \geq 3$ : range $[3, + \infty [$ .
Common mistake: Giving $R$ .
Show that $f (x) = \frac{x}{x ^{2} + 1}$ is odd.
Show answer
$f (- x) = \frac{- x}{x ^{2} + 1} = - f (x)$ , and $R$ is symmetric.
Common mistake: Testing a single value only.
Decompose $\frac{x ^{2} + 1}{x ^{2} - 1}$ .
Show answer
Divide: $1 + \frac{2}{x ^{2} - 1} = 1 + \frac{1}{x - 1} - \frac{1}{x + 1}$ .
Common mistake: Decomposing without dividing first.
Find $f^{- 1}$ for $f (x) = x - 1$ , $x \geq 1$ .
Show answer
$y = x - 1 \Rightarrow x = y^{2} + 1$ ; $f^{- 1} (x) = x^{2} + 1$ for $x \geq 0$ .
Common mistake: Forgetting to restrict the domain of the inverse.
Decompose $\frac{5 x + 1}{( x - 1 ) ( x + 2 )}$ .
Show answer
$\frac{2}{x - 1} + \frac{3}{x + 2}$ .
Common mistake: Substituting values that don't eliminate a constant.

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Find the domain of $g (x) = \frac{1}{9 - x ^{2}}$ .	a. $[2, 5 [\cup] 5, + \infty [$
2. Decompose $\frac{x + 3}{x ^{2} - x - 2}$ into partial fractions.	b. $R ∖ {- 1, 4}$
3. Find the domain of $h (x) = \frac{x + 1}{x ^{2} - 3 x - 4}$ .	c. $\frac{5}{3 ( x - 2 )} - \frac{2}{3 ( x + 1 )}$
4. Find the domain of $f (x) = \frac{x - 2}{x - 5}$ .	d. $] - 3, 3 [$

Show answer

1 → d ( $] - 3, 3 [$ ) · 2 → c ( $\frac{5}{3 ( x - 2 )} - \frac{2}{3 ( x + 1 )}$ ) · 3 → b ( $R ∖ {- 1, 4}$ ) · 4 → a ( $[2, 5 [\cup] 5, + \infty [$ )

Step 1. The root needs $9 - x^{2} \geq 0$ ; it is in a denominator, so $9 - x^{2} > 0$ .
Step 2. $x^{2} < 9$ : $- 3 < x < 3$ .
Step 1. $x^{2} - x - 2 = (x - 2) (x + 1)$ ; $x + 3 = A (x + 1) + B (x - 2)$ .
Step 2. $x = 2$ : $5 = 3 A$ , $A = \frac{5}{3}$ . $x = - 1$ : $2 = - 3 B$ , $B = - \frac{2}{3}$ .
Step 1. $x^{2} - 3 x - 4 = (x - 4) (x + 1)$ .
Step 2. Exclude both zeros of the denominator, even though $x + 1$ also appears above.
Step 1. Square root: $x - 2 \geq 0$ , so $x \geq 2$ .
Step 2. Denominator: $x \neq = 5$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$[0, + \infty [$ · even · $[4, + \infty [$ · odd
1. The function $f (x) = x^{4} - 3 x^{2} + 2$ is .
2. The function $f (x) = x^{3} - 4 x$ is .
3. $f (x) = x$ and $g (x) = x - 4$ . The domain of $f \circ g$ is .
Show answer
1: even · 2: odd · 3: $[4, + \infty [$
- Step 1. $f (- x) = x^{4} - 3 x^{2} + 2 = f (x)$ (even powers only).
  Step 2. So $f$ is even.
- Step 1. $f (- x) = - x^{3} + 4 x = - (x^{3} - 4 x)$ .
  Step 2. $f (- x) = - f (x)$ : odd.
- Step 1. $(f \circ g) (x) = x - 4$ .
  Step 2. $x - 4 \geq 0$ : $x \geq 4$ .

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. $f (x) = 2 x + 1$ . Solve $f (f (x)) = 11$ .	a. $] - \infty, - 2] \cup [2, + \infty [$
2. $f (x) = x$ and $g (x) = x^{2} - 4$ . Find the domain of $f \circ g$ .	b. $g (x) = 2 x + 1$
3. Decompose $\frac{5 x + 1}{( x - 1 ) ( x + 2 )}$ into partial fractions.	c. $x = 2$
4. $f (x) = 3 x - 2$ and $(f \circ g) (x) = 6 x + 1$ . Find $g (x)$ .	d. $\frac{2}{x - 1} + \frac{3}{x + 2}$

Show answer

1 → c ( $x = 2$ ) · 2 → a ( $] - \infty, - 2] \cup [2, + \infty [$ ) · 3 → d ( $\frac{2}{x - 1} + \frac{3}{x + 2}$ ) · 4 → b ( $g (x) = 2 x + 1$ )

Step 1. $f (f (x)) = 2 (2 x + 1) + 1 = 4 x + 3$ .
Step 2. $4 x + 3 = 11$ , so $x = 2$ .
Step 1. $(f \circ g) (x) = x^{2} - 4$ .
Step 2. Need $x^{2} - 4 \geq 0$ : $x \leq - 2$ or $x \geq 2$ .
Step 1. $5 x + 1 = A (x + 2) + B (x - 1)$ .
Step 2. $x = 1$ : $6 = 3 A$ , $A = 2$ . $x = - 2$ : $- 9 = - 3 B$ , $B = 3$ .
Step 1. $f (g (x)) = 3 g (x) - 2 = 6 x + 1$ .
Step 2. $3 g (x) = 6 x + 3$ , so $g (x) = 2 x + 1$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$] - \infty, - 3 [\cup [1, + \infty [$ · $2 x^{2} - 1$ · $f^{- 1} (x) = 3 x - 4$ · $2 x^{2} - 3$
1. $f (x) = x^{2} + 1$ and $g (x) = 2 x - 3$ . Find $(g \circ f) (x)$ . →
2. Find the domain of $\frac{x - 1}{x + 3}$ . →
3. Find the inverse of $f (x) = \frac{x + 4}{3}$ . →
Show answer
1: $2 x^{2} - 1$ · 2: $] - \infty, - 3 [\cup [1, + \infty [$ · 3: $f^{- 1} (x) = 3 x - 4$
- Step 1. $(g \circ f) (x) = g (f (x)) = 2 (x^{2} + 1) - 3$ .
  Step 2. $= 2 x^{2} - 1$ .
- Step 1. Need $\frac{x - 1}{x + 3} \geq 0$ , with $x \neq = - 3$ .
  Step 2. Sign table: positive for $x < - 3$ and for $x \geq 1$ .
- Step 1. $y = \frac{x + 4}{3}$ , so $3 y = x + 4$ and $x = 3 y - 4$ .
  Step 2. Swap the names: $f^{- 1} (x) = 3 x - 4$ .

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