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 ) = 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)

8. Rational Expressions and Functions

Simplifying Rational Expressions

Intermediate Algebra

8. Rational Expressions and Functions

Simplifying Rational Expressions

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Multiple Choice

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

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

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

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

$(- \infty, \infty)$

Verified step by step guidance

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 $(-\infty, 1) \cup (1, 4) \cup (4, \infty)$.

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