Determine the domain of the function f ( x ) f\left(x\right) . f ( x ) = 5 x f\left(x\right)=\frac{5}{x}

( − ∞ , 0 ) ∪ ( 0 , ∞ ) \left(-\infty,0\right)\cup\left(0,\infty\right)

( − ∞ , 0 ] ∪ [ 0 , ∞ ) \left(-\infty,0\right\rbrack\cup\left\lbrack0,\infty\right)

( − ∞ , ∞ ) \left(-\infty,\infty\right)

( 0 , ∞ ) \left(0,\infty\right)

Determine the domain of the function f ( x ) f\left(x\right) . f ( x ) = 7 x − 3 f\left(x\right)=\frac{7}{x-3}

( − ∞ , 3 ) ∪ ( 3 , ∞ ) \left(-\infty,3\right)\cup\left(3,\infty\right)

( − ∞ , ∞ ) \left(-\infty,\infty\right)

( 3 , ∞ ) \left(3,\infty\right)

( − ∞ , 3 ] ∪ [ 3 , ∞ ) \left(-\infty,3\right\rbrack\cup\left\lbrack3,\infty\right)

Determine the domain of the function g ( x ) g\left(x\right) . g ( x ) = x + 1 x 2 − 9 g\left(x\right)=\frac{x+1}{x^2-9}

( − ∞ , − 3 ) ∪ ( − 3 , 3 ) ∪ ( 3 , ∞ ) \left(-\infty,-3\right)\cup\left(-3,3\right)\cup\left(3,\infty\right)

( − ∞ , ∞ ) \left(-\infty,\right.\left.\infty\right)

( − ∞ , − 3 ] ∪ [ 3 , ∞ ) \left(-\infty,-3\right\rbrack\cup\left\lbrack3,\infty\right)

( − ∞ , − 3 ) ∪ ( 3 , ∞ ) \left(-\infty,-3\right)\cup\left(3,\infty\right)

Determine the domain of the function h ( x ) h\left(x\right) . h ( x ) = x 2 − 2 x − 8 x 2 − 5 x + 4 h\left(x\right)=\frac{x^2-2x-8}{x^2-5x+4}

( − ∞ , 1 ) ∪ ( 1 , 4 ) ∪ ( 4 , ∞ ) \left(-\infty,1\right)\cup\left(1,4\right)\cup\left(4,\infty\right)

( − ∞ , 4 ) ∪ ( 4 , ∞ ) \left(-\infty,4\right)\cup\left(4,\infty\right)

( − ∞ , 4 ] ∪ [ 4 , ∞ ) \left(-\infty,4\right\rbrack\cup\left\lbrack4,\infty\right)

( − ∞ , ∞ ) \left(-\infty,\infty\right)

Simplify the rational expressions below: x 2 − 9 x 2 − 3 x \frac{x^2-9}{x^2-3x}

x − 3 x \frac{x-3}{x} x x − 3

Simplify the rational expressions below: x 2 − 4 x x 2 − 2 x − 8 x \frac{x^2-4x}{x^2-2x-8x}

x − 4 x − 10 \frac{x-4}{x-10} x − 10 x − 4

x + 4 x + 10 \frac{x+4}{x+10} x + 10 x + 4

x − 2 x − 5 \frac{x-2}{x-5} x − 5 x − 2

x + 2 x + 5 \frac{x+2}{x+5} x + 5 x + 2

Simplify the rational expressions below: x 2 + 5 x + 6 x 2 + 7 x + 10 \frac{x^2+5x+6}{x^2+7x+10}

x + 3 x + 5 \frac{x+3}{x+5} x + 5 x + 3

x + 3 x + 2 \frac{x+3}{x+2} x + 2 x + 3

x + 2 x + 2 \frac{x+2}{x+2} x + 2 x + 2

How do you evaluate a rational function at a specific value?

To evaluate a rational function at a specific value, substitute the value into the function's expression and simplify. For example, given f ( x )= x-1 2x-6 , to find f(2) , replace x with 2: 2-1 2 imes 2 - 6 = 1 4-6 = 1 -2 = -2 . This method requires careful substitution and simplification, ensuring the denominator is not zero at the chosen value.

8. Rational Expressions and Functions

Simplifying Rational Expressions

8. Rational Expressions and Functions

Simplifying Rational Expressions: Videos & Practice Problems

Video Lessons

Topic summary

Rational expressions are quotients of polynomials, where the denominator cannot be zero to avoid undefined values. Simplifying these expressions involves factoring the numerator and denominator completely, then canceling common factors, similar to simplifying rational numbers. For example, factoring $3x^2 - 15x$ by extracting the greatest common factor reveals common binomial factors that can be canceled. Understanding domain restrictions and simplification techniques is essential for working with rational functions, which are defined by rational expressions. This knowledge builds foundational skills in algebra involving polynomials, terms, coefficients, and exponents.

concept

Intro to Rational Expressions and Functions

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Intro to Rational Expressions and Functions Video Summary

Rational expressions are algebraic fractions where both the numerator and denominator are polynomials. Typically, the numerator is represented as p and the denominator as q, forming the rational expression $\frac{p}{q}$. For example, the expression $\frac{4x}{x - 2}$ is a rational expression because it is a quotient of two polynomials. A crucial rule for rational expressions is that the denominator cannot be zero, as division by zero is undefined in mathematics. This means any value of the variable that makes the denominator zero must be excluded from the domain.

Building on this, a rational function is defined by a rational expression expressed in function notation, such as $f(x) = \frac{p(x)}{q(x)}$. The domain of a rational function includes all real numbers except those that make the denominator zero. To find the domain, set the denominator equal to zero and solve for the variable. The solutions represent values that must be excluded from the domain.

For instance, consider the rational function $f(x) = \frac{x - 1}{2x - 6}$. To find its domain, set the denominator equal to zero:

\[2x - 6 = 0\]

Solving for $x$:

\[2x = 6 \\x = \frac{6}{2} = 3\]

Since $x = 3$ makes the denominator zero, it is excluded from the domain. Thus, the domain in set-builder notation is:

\[\{ x \in \mathbb{R} \mid x \neq 3 \}\]

In interval notation, this is expressed as:

\[(-\infty, 3) \cup (3, \infty)\]

Evaluating the function at a specific value involves substituting that value into the expression. For example, to find $f(2)$:

\[f(2) = \frac{2 - 1}{2(2) - 6} = \frac{1}{4 - 6} = \frac{1}{-2} = -\frac{1}{2}\]

Understanding rational expressions and functions involves recognizing their structure as quotients of polynomials, identifying restrictions on the domain due to zero denominators, and performing evaluations by substitution. Mastery of these concepts is essential for solving algebraic problems involving rational functions and for further studies in calculus and advanced mathematics.

Problem

Determine the domain of the function $f (x)$ .
$f (x) = \frac{5}{x}$

$open paren negative infinity comma 0 close paren union open paren 0 comma infinity close paren$

$open paren negative infinity comma 0 close bracket union open bracket 0 comma infinity close paren$

$(- \infty, \infty)$

$open paren 0 comma infinity close paren$

Problem

Determine the domain of the function $f (x)$ .
$f (x) = \frac{7}{x - 3}$

$(- \infty, \infty)$

$open paren 3 comma infinity close paren$

$open paren negative infinity comma 3 close paren union open paren 3 comma infinity close paren$

$open paren negative infinity comma 3 close bracket union open bracket 3 comma infinity close paren$

Problem

Determine the domain of the function $g (x)$ .
$g (x) = \frac{x + 1}{x^{2} - 9}$

$(- \infty, \infty)$

$(- \infty, - 3] \cup [3, \infty)$

$(- \infty, - 3) \cup (3, \infty)$

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Problem

Determine the domain of the function $h (x)$ .
$h (x) = \frac{x^{2} - 2 x - 8}{x^{2} - 5 x + 4}$

$open paren negative infinity comma 4 close paren union open paren 4 comma infinity close paren$

$open paren negative infinity comma 1 close paren union open paren 1 comma 4 close paren union open paren 4 comma infinity close paren$

$open paren negative infinity comma 4 close bracket union open bracket 4 comma infinity close paren$

$(- \infty, \infty)$

Problem

Given the function below, evaluate $f of 2$
$f (x) = \frac{3}{11 - x}$

$3$

$\frac{1}{3}$

$3 over 13$

$\frac{2}{3}$

Problem

Given the function below, evaluate $f of 0$
$f (x) = \frac{3 x + 4}{x^{2} + 7 x + 20}$

$\frac{1}{5}$

$\frac{4}{5}$

$\infty$

$7 over 20$

concept

Simplifying Rational Expressions

Video duration:

Simplifying Rational Expressions Video Summary

Rational expressions are fractions where the numerator and denominator are polynomials. Simplifying rational expressions involves factoring both the numerator and denominator completely and then canceling any common factors, much like simplifying numerical fractions. For example, to simplify a fraction like $\frac{28}{35}$, you factor both numbers: \$28 = 7 \times 4$ and $35 = 7 \times 5$. Since 7 is a common factor, it can be canceled, leaving $\frac{4}{5}\( as the simplified form.

When simplifying rational expressions with variables, the process is similar. Consider the expression \)\frac{28x^3}{35x^5}$. First, factor the constants and express the variables as repeated factors: $28 = 7 \times 4$, $35 = 7 \times 5$, $x^3 = x \times x \times x$, and $x^5 = x \times x \times x \times x \times x$. Cancel the common factors of 7 and three $x$'s from numerator and denominator. This leaves $\frac{4}{5x^2}$ as the simplified expression, since two $x$'s remain in the denominator, which is written as $x^2\(.

In cases where expressions are already factored, simplification is straightforward. For instance, given \)\frac{(x+2)(x-2)}{(x+6)(x+2)}$, the common factor $(x+2)$ cancels, resulting in $\frac{x-2}{x+6}\(. This highlights the importance of recognizing and canceling common binomial factors.

Sometimes, expressions require factoring before simplification. For example, in \)\frac{x-5}{3x^2 - 15x}$, the denominator can be factored by extracting the greatest common factor (GCF). Since $3x$ is the GCF of $3x^2$ and $15x$, factor it out: $3x^2 - 15x = 3x(x - 5)$. Now the expression becomes $\frac{x-5}{3x(x-5)}$. The common factor $(x-5)$ cancels, leaving the simplified form $\frac{1}{3x}$.

Understanding how to factor polynomials and identify common factors is essential for simplifying rational expressions. This process not only reduces expressions to their simplest form but also prepares you for more advanced algebraic operations. Remember, the key steps are to factor completely, cancel common factors, and rewrite the expression in simplest terms.

Problem

Simplify the rational expressions below:
$6 x over 12 x$

$2x$

$2$

$\frac12x$

$\frac12$

Problem

Simplify the rational expressions below:
$\frac{x^{2} - 9}{x^{2} - 3 x}$

$3$

$\frac{x}{x+3}$

$\frac{x+3}{x}$

$\frac{x-3}{x}$

Problem

Simplify the rational expressions below:
$\frac{x^{2} - 4 x}{x^{2} - 2 x - 8 x}$

$\frac{x-4}{x-10}$

$\frac{x+4}{x+10}$

$\frac{x-2}{x-5}$

$\frac{x+2}{x+5}$

example

Simplifying Rational Expressions Example 1

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Simplifying Rational Expressions Example 1 Video Summary

Understanding how to write equivalent forms of rational expressions is a fundamental skill in algebra that enhances problem-solving flexibility. Consider the rational expression $$\frac{-x^2 + 4x}{x^2 - 2x}$$ . One effective method to find equivalent expressions is through factoring both the numerator and denominator.

Start by factoring a negative $$x$$ from the numerator. Since $$-x \times x = -x^2$$ and $$-x \times (-4) = 4x$$ , the numerator factors to $$-x(x - 4)$$ . Similarly, factor a negative $$x$$ from the denominator: $$-x \times (-x) = x^2$$ and $$-x \times 2 = -2x$$ , so the denominator becomes $$-x(-x + 2)$$ . Canceling the common factor $$-x$$ leaves the equivalent expression $$\frac{x - 4}{2 - x}$$ , noting that $$-x + 2$$ is rewritten as $\$2 - x\($ for clarity.

Another approach involves factoring a positive $\)x$$ from the denominator instead. Keeping the numerator as $$-x(x - 4)$$ , factor the denominator as $$x(x - 2)$$ . Canceling the common factor $$x$$ results in $$\frac{- (x - 4)}{x - 2}$$ , which simplifies to $$\frac{4 - x}{x - 2}\($ by distributing the negative sign in the numerator. This form highlights how factoring choices affect the appearance but not the value of the expression.

A third method to generate an equivalent expression is by multiplying both numerator and denominator by the same nonzero constant, such as 2. Multiplying yields $\)\frac{2(-x^2 + 4x)}{2(x^2 - 2x)} = \frac{-2x^2 + 8x}{2x^2 - 4x}$$ . This operation preserves equivalence because multiplying by $$\frac{2}{2}$$ is effectively multiplying by 1. Rearranging terms for clarity, the expression can be written as $$\frac{8x - 2x^2}{2x^2 - 4x}$$ . This technique demonstrates that scaling both parts of a rational expression by the same factor produces infinitely many equivalent forms.

Mastering these strategies—factoring with different signs, rearranging terms, and multiplying by constants—enables the creation of diverse equivalent expressions. This flexibility is crucial for simplifying complex algebraic fractions, solving equations, and understanding the underlying structure of rational expressions.

example

Simplifying Rational Expressions Example 2

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Simplifying Rational Expressions Example 2 Video Summary

To determine which rational expression equals negative five, we simplify each option step-by-step. Starting with the first expression, factor the numerator as \$2 \times 5 \times x$ and the denominator as $2 \times x$. Canceling the common factors $2$ and $x$ leaves $5$, which is not equal to $-5\(, so this option is incorrect.

Next, consider the expression \)\frac{25 - 5x}{x}$. Recognize that $25$ is $5^2$, and factor out $5$ from the numerator to get $5(5 - x)$. Since the denominator is $x$, no further simplification or cancellation is possible, and the expression does not simplify to $-5\(.

For the third expression, \)\frac{-15x}{-3x}$, bring the negative signs to the numerator and denominator, which cancel each other out. Factor $15$ as $3 \times 5$, so the expression becomes $\frac{3 \times 5 \times x}{3 \times x}$. Canceling $3$ and $x$ leaves $5$, which again is not equal to $-5\(.

Finally, examine the expression \)\frac{20 - 5x}{x - 4}$. Factor $5$ from the numerator to get $5(4 - x)$. Notice that $4 - x$ can be rewritten as $-(x - 4)$ because distributing the negative sign yields $-x + 4$, which is equivalent to $4 - x$. This allows rewriting the numerator as $5 \times [-(x - 4)] = -5(x - 4)$. The $(x - 4)$ terms in numerator and denominator cancel, leaving $-5\(. Therefore, this rational expression simplifies exactly to negative five.

Through factoring, recognizing equivalent expressions, and canceling common terms, the expression \)\frac{20 - 5x}{x - 4}$ is confirmed to be equal to $-5$. This process highlights the importance of factoring and understanding how to manipulate expressions to reveal equivalences in rational expressions.

Problem

Simplify the rational expressions below:
$\frac{x^{2} + 5 x + 6}{x^{2} + 7 x + 10}$

$\frac{x+3}{x+2}$

$\frac{x+3}{x+5}$

$\frac{x+2}{x+2}$

$1$

Here’s what students ask on this topic:

A rational expression is a quotient of two polynomials, where the numerator and denominator are polynomials, and the denominator cannot be zero. This is similar to a rational number, which is a quotient of two integers. The key difference is that rational expressions involve variables and polynomials, while rational numbers are fixed fractions of integers. For example, $\frac{4x}{x-2}$ is a rational expression. Understanding that the denominator cannot be zero is crucial because division by zero is undefined. This restriction also affects the domain of rational functions, which are functions defined by rational expressions.

To find the domain of a rational function, you need to identify all values of the variable that make the denominator zero, because division by zero is undefined. You do this by setting the denominator equal to zero and solving for the variable. For example, if the function is $f (x)= \frac{x-1}{2x-6}$ , set $2x-6=0$ . Solving gives $x=3$ . Therefore, the domain is all real numbers except $3$ . In interval notation, this is $(-2E2,3)2E2, 2E2)$ . This ensures the function is defined for all allowed inputs.

Simplifying a rational expression involves three main steps: First, factor the numerator and denominator completely into their prime factors or polynomial factors. Second, identify and cancel any common factors that appear in both numerator and denominator. Third, write the simplified expression with the remaining factors. For example, to simplify $\frac{28x^3}{35x^5}$ , factor as $\frac{7 imes 4 imes x imes x imes x}{7 imes 5 imes x imes x imes x imes x imes x}$ . Cancel the common factor 7 and three x's to get $\frac{4}{5x^2}$ . This process is similar to simplifying numeric fractions but involves variables and exponents.

The denominator of a rational expression cannot be zero because division by zero is undefined in mathematics. If the denominator equals zero for some value of the variable, the expression or function is not defined at that point. This leads to restrictions on the domain of rational functions. For example, in the expression $\frac{x}{x-2}$ , if $x=2$ , the denominator becomes zero, making the expression undefined. Therefore, we exclude such values from the domain to ensure the function is valid and meaningful.

To evaluate a rational function at a specific value, substitute the value into the function's expression and simplify. For example, given $f (x)= \frac{x-1}{2x-6}$ , to find $f(2)$ , replace $x$ with 2: $\frac{2-1}{2 imes 2 - 6} = \frac{1}{4-6} = \frac{1}{-2} = -2$ . This method requires careful substitution and simplification, ensuring the denominator is not zero at the chosen value.

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Simplifying Rational Expressions: Videos & Practice Problems

Intro to Rational Expressions and Functions

Intro to Rational Expressions and Functions Video Summary

Determine the domain of the function $f (x)$ .
$f (x) = \frac{5}{x}$

Determine the domain of the function $f (x)$ .
$f (x) = \frac{7}{x - 3}$

Determine the domain of the function $g (x)$ .
$g (x) = \frac{x + 1}{x^{2} - 9}$

Determine the domain of the function $h (x)$ .
$h (x) = \frac{x^{2} - 2 x - 8}{x^{2} - 5 x + 4}$

Given the function below, evaluate $f of 2$
$f (x) = \frac{3}{11 - x}$

Given the function below, evaluate $f of 0$
$f (x) = \frac{3 x + 4}{x^{2} + 7 x + 20}$

Simplifying Rational Expressions

Simplifying Rational Expressions Video Summary

Simplify the rational expressions below:
$6 x over 12 x$

Simplify the rational expressions below:
$\frac{x^{2} - 9}{x^{2} - 3 x}$

Simplify the rational expressions below:
$\frac{x^{2} - 4 x}{x^{2} - 2 x - 8 x}$

Simplifying Rational Expressions Example 1

Simplifying Rational Expressions Example 1 Video Summary

Simplifying Rational Expressions Example 2

Simplifying Rational Expressions Example 2 Video Summary

Simplify the rational expressions below:
$\frac{x^{2} + 5 x + 6}{x^{2} + 7 x + 10}$

Here’s what students ask on this topic:

What is a rational expression and how is it different from a rational number?

How do you find the domain of a rational function?

What are the steps to simplify a rational expression?

Why can't the denominator of a rational expression be zero?

How do you evaluate a rational function at a specific value?

Your Intermediate Algebra tutors

Simplifying Rational Expressions: Videos & Practice Problems

Intro to Rational Expressions and Functions

Intro to Rational Expressions and Functions Video Summary

Determine the domain of the function f(x)f\left(x\right).f(x)=5xf\left(x\right)=\frac{5}{x}

Determine the domain of the function f(x)f\left(x\right).f(x)=7x−3f\left(x\right)=\frac{7}{x-3}

Determine the domain of the function g(x)g\left(x\right).g(x)=x+1x2−9g\left(x\right)=\frac{x+1}{x^2-9}

Determine the domain of the function h(x)h\left(x\right).h(x)=x2−2x−8x2−5x+4h\left(x\right)=\frac{x^2-2x-8}{x^2-5x+4}

Given the function below, evaluate f(2)f\left(2\right)f(x)=311−xf\left(x\right)=\frac{3}{11-x}

Given the function below, evaluate f(0)f\left(0\right)f(x)=3x+4x2+7x+20f\left(x\right)=\frac{3x+4}{x^2+7x+20}

Simplifying Rational Expressions

Simplifying Rational Expressions Video Summary

Simplify the rational expressions below:6x12x\frac{6x}{12x}

Simplify the rational expressions below:x2−9x2−3x\frac{x^2-9}{x^2-3x}

Simplify the rational expressions below:x2−4xx2−2x−8x\frac{x^2-4x}{x^2-2x-8x}

Simplifying Rational Expressions Example 1

Simplifying Rational Expressions Example 1 Video Summary

Simplifying Rational Expressions Example 2

Simplifying Rational Expressions Example 2 Video Summary

Simplify the rational expressions below:x2+5x+6x2+7x+10\frac{x^2+5x+6}{x^2+7x+10}

Here’s what students ask on this topic:

What is a rational expression and how is it different from a rational number?

What is a rational expression and how is it different from a rational number?

How do you find the domain of a rational function?

How do you find the domain of a rational function?

What are the steps to simplify a rational expression?

What are the steps to simplify a rational expression?

Why can't the denominator of a rational expression be zero?

Why can't the denominator of a rational expression be zero?

How do you evaluate a rational function at a specific value?

How do you evaluate a rational function at a specific value?

Your Intermediate Algebra tutors

Determine the domain of the function $f (x)$ .
$f (x) = \frac{5}{x}$

Determine the domain of the function $f (x)$ .
$f (x) = \frac{7}{x - 3}$

Determine the domain of the function $g (x)$ .
$g (x) = \frac{x + 1}{x^{2} - 9}$

Determine the domain of the function $h (x)$ .
$h (x) = \frac{x^{2} - 2 x - 8}{x^{2} - 5 x + 4}$

Given the function below, evaluate $f of 2$
$f (x) = \frac{3}{11 - x}$

Given the function below, evaluate $f of 0$
$f (x) = \frac{3 x + 4}{x^{2} + 7 x + 20}$

Simplify the rational expressions below:
$6 x over 12 x$

Simplify the rational expressions below:
$\frac{x^{2} - 9}{x^{2} - 3 x}$

Simplify the rational expressions below:
$\frac{x^{2} - 4 x}{x^{2} - 2 x - 8 x}$

Simplify the rational expressions below:
$\frac{x^{2} + 5 x + 6}{x^{2} + 7 x + 10}$