Find the domain of the rational function. Then, write it in lowest terms.f(x)=x2+9x−3f\left(x\right)=\frac{x^2+9}{x-3}f(x)=x−3x2+9

{x∣x≠3}, f(x)=x2+9x−3f\left(x\right)=\frac{$$x^2+9$$}{x-3}f(x)=x−3x2+9

Find the domain of the rational function. Then, write it in lowest terms. f ( x ) = x 2 + 9 x − 3 f\left(x\right)=\frac{x^2+9}{x-3} f ( x ) = x − 3 x 2 + 9

Hey, everyone. Let's take a look at this practice problem. Here, we want to find the domain of the rational function and then write it in lowest terms. So let's go ahead and start by finding the domain. Now when we find the domain, we want to take our denominator and set it equal to 0. So here doing that, I can go ahead and solve for x by simply adding adding 3 to both sides. Now that will cancel over here, leaving me with x is equal to 3, and this represents my domain restriction. So I know that my domain is now x such that x x cannot be equal to 3 because that would make my denominator 0. So this is our domain. Let's go ahead and write this in lowest terms. So taking another look at my function here, f of x is equal to x squared plus 9 divided by x minus 3. Remember, we want to fully factor both the top and the bottom of our fraction. But looking at the top here, x squared plus 9 is actually not going to factor any further. It's just going to remain x squared plus 9. Now our denominator also cannot be factored any further. It's just x minus 3. So this is actually already in lowest terms, x squared plus 9 divided by x minus 3. So I don't need to do anything. I don't need to cancel anything out with my function. And this is my function in lowest terms. Thanks for watching, and I'll see you in the next video.

Find the domain of the rational function. Then, write it in lowest terms.f(x)=x2+9x−3f\left(x\right)=\frac{$$x^2+9$$}{x-3}f(x)=x−3x2+9

{x∣x≠3}, f(x)=x2+9x−3f\left(x\right)=\frac{$$x^2+9$$}{x-3}f(x)=x−3x2+9

{x∣x≠0}, f(x)=1x−3f\left(x\right)=\frac{1}{x-3}f(x)=x−31

{x∣x≠−3}, f(x)=x2+9x−3f\left(x\right)=\frac{$$x^2+9$$}{x-3}f(x)=x−3x2+9

{x∣x≠3}, f(x)=x+3f\left(x\right)=x+3f(x)=x+3

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5. Rational Functions

Introduction to Rational Functions

Precalculus

5. Rational Functions

Introduction to Rational Functions

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Multiple Choice

Find the domain of the rational function. Then, write it in lowest terms.
$f$\left$(x$\right$)=$\frac{x^2+9}{x-3}$$

{x∣x≠0}, $f$\left$(x$\right$)=$\frac{1}{x-3}$$

{x∣x≠3}, $f$\left$(x$\right$)=$\frac{x^2+9}{x-3}$$

{x∣x≠−3}, $f$\left$(x$\right$)=$\frac{x^2+9}{x-3}$$

{x∣x≠3}, $f$\left$(x$\right$)=x+3$

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Verified step by step guidance

Identify the rational function given: $ f(x) = \frac{x^2 + 9}{x - 3} $.

Determine the domain of the function by identifying values of $ x $ that make the denominator zero. Set the denominator equal to zero: $ x - 3 = 0 $. Solve for $ x $ to find the value that is not in the domain.

The solution to $ x - 3 = 0 $ is $ x = 3 $. Therefore, the domain of the function is all real numbers except $ x = 3 $. In set notation, this is $ \{ x \mid x \neq 3 \} $.

Simplify the rational function if possible. Check if the numerator $ x^2 + 9 $ can be factored and if any common factors exist with the denominator $ x - 3 $.

Since $ x^2 + 9 $ cannot be factored further and has no common factors with $ x - 3 $, the function is already in its lowest terms. Thus, the simplified form of the function remains $ f(x) = \frac{x^2 + 9}{x - 3} $.