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}

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

this problem, we are told to determine the domain of the function h of x. And notice we have h of x as this rational function. Now right off the bat, this rational function actually looks a bit complicated, but there is a straightforward way we can go about handling this. The thing we need to remember for any rational function is that the denominator cannot be equal to zero, because that would be dividing by zero, which is not allowed. So these are going to be the restrictions on our domain. So what I'm looking for for the domain is all the allowed x values, and that's what I'm trying to find. So what I can say is that focusing strictly on my denominator here, x squared minus 5x plus four cannot equal zero. So what I need to do is think of how I could go about solving this equation. And it is a bit tricky just initially looking at this, but my strategy is going to be factoring. So So I notice that I have a quadratic right here, where the leading coefficient is one on this x squared. So because of that, all I need to do is find a pair of numbers that multiply to give me positive four, this last term, and then add to give me negative five. Now it's gonna be hard to just list out every single possible pair here, but there is a couple of numbers that come to mind. Think about this, negative one and negative four. If you think about this, multiplying them together would cancel those negatives, giving me positive four. And adding these numbers would give me negative one plus negative four, which I know is negative five. So this is the pair of numbers I'm looking for. So what that means is this entire equation here, the x squared minus five x plus four, this will factor into x minus one times x minus four. And keep in mind, this whole thing cannot be equal to zero. Now I want to solve for x, and I'm gonna get two different equations that come out of this. I'll get that x minus one cannot equal zero from this side. Then I'll also get that x minus four cannot equal zero from this piece. So now that we have this, we can go ahead and solve each of these equations. So I'll add one to both sides of this equation, canceling the ones there, giving me that x cannot equal positive one. Now I'll also add four on both sides of the other equation to get the force to cancel there, leaving me with x cannot equal positive four. So we have that x cannot equal one, and x cannot equal four. And this tells us exactly the restrictions on our domain. So our domain can be any real number except for one and four. Now there's different ways that I could write this, but what I'm going to do is use interval notation. So if we're using interval notation, if I want to write out the domain here, well that's going to be every number from negative infinity all the way up to one. Then we're going to have every number between one all the way up to four, and then it's going to be every number from four all the way up to infinity, but notice that I'm putting parentheses around everything. That's because one and four are not included, and we never include infinity of any kind, negative or positive. So because of this, I need to put parentheses over everything, and this is just telling me, hey, we can have any real number here between negative infinity until we get to one. One is the number we cannot have. We can have any number between one and four, but once we get to four, we can't have that number either, but then we can have any number from four up. So this tells me that my domain is any real numbers except for one and four. So hope you found this video helpful, and I'll see you all in the next one.

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

(−∞,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)

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1. Review of Real Numbers2h 43m

Simplifying Fractions
30m

Evaluating Exponents
29m

Evaluating Expressions
15m

Translating Phrases to Expressions
16m

Translating Sentences to Equations
9m

The Distributive Property
17m

Simplifying Expressions
44m

2. Linear Equations and Inequalities5h 35m

Solving Linear Equations
2h 23m

Linear Inequalities in One Variable
46m

Set Operations and Compound Inequalities
1h 7m

Absolute Value Equations
38m

Absolute Value Inequalities
39m

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Introduction to Problem Solving
37m

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25m

Percent Problem Solving
59m

Mixture Problem Solving
43m

4. Graphs and Functions5h 12m

The Rectangular Coordinate System
44m

Graph Linear Equations in Two Variables
24m

Graph Linear Equations Using Intercepts
23m

Slope of a Line
44m

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38m

Point Slope Form
22m

Linear Inequalities in Two Variables
28m

Introduction to Relations and Functions
53m

Function Notation
15m

Composition of Functions
17m

5. Systems of Linear Equations1h 53m

Solving Systems of Linear Equations by Graphing
29m

Solving Systems of Linear Equations by Substitution
15m

Solving Systems of Linear Equations by Elimination
45m

Systems of Linear Inequalities
23m

6. Exponents, Polynomials, and Polynomial Functions3h 17m

The Product Rule
9m

The Quotient Rule
10m

Intro to the Power Rules
17m

The Power of a Quotient Rule
13m

Negative Exponents
24m

Simplifying Exponential Expressions Using All Exponent Rules
6m

Intro to Polynomials
21m

Adding and Subtracting Polynomials
20m

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38m

Special Products
34m

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Factoring by Greatest Common Factor & Grouping
38m

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11m

Factoring Trinomials of the Form ax² + bx + c
37m

Factoring Special Products
41m

Quadratic Equations & Applications
41m

8. Rational Expressions and Functions3h 44m

Simplifying Rational Expressions
42m

Multiplying and Dividing Rational Expressions
25m

Adding and Subtracting Rational Expressions with Common Denominators
19m

Least Common Denominators
32m

Adding and Subtracting Rational Expressions with Different Denominators
32m

Rational Equations
44m

Direct & Inverse Variation
27m

9. Roots, Radicals, and Complex Numbers3h 57m

Introduction to Radical Expressions
47m

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47m

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19m

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53m

Introduction to Complex Numbers
18m

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51m

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The Square Root Property
25m

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26m

The Quadratic Formula
40m

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1h 28m

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25m

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35m

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33m

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54m

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32m

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35m

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35m

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41m

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Introduction to Sequences
40m

Arithmetic Sequences
27m

Geometric Sequences
29m

Factorials
13m

8. Rational Expressions and Functions

Simplifying Rational Expressions

Intermediate Algebra

8. Rational Expressions and Functions

Simplifying Rational Expressions

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

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

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

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

$(- \infty, \infty)$

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Problema precedenteProssimo problema

Guida verificata passo dopo passo

Identify the function given: $h\left(x\right) = \frac{x^2 - 2x - 8}{x^2 - 5x + 4}$. To find the domain, we need to determine where this function is defined, which means the denominator cannot be zero.

Set the denominator equal to zero to find the values of $x$ that are not allowed: $x^2 - 5x + 4 = 0$.

Factor the quadratic in the denominator: $x^2 - 5x + 4 = (x - 1)(x - 4)$.

Solve each factor equal to zero to find the excluded values: $x - 1 = 0$ gives $x = 1$, and $x - 4 = 0$ gives $x = 4$. These values make the denominator zero, so they are not in the domain.

Write the domain as all real numbers except $x = 1$ and $x = 4$, which can be expressed in interval notation as $(-\infty, 1) \cup (1, 4) \cup (4, \infty)$.