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}

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

this problem, we are asked to determine the domain of the function g of x. So this is the function we have. It's a rational function. And I know for rational functions, the denominator cannot be equal to zero. So we need to figure out which x values we are not allowed to have here. What would be the restrictions on our domain? So what I'm going to do is take this denominator, x squared minus nine, and I know that this cannot be equal to zero. So what I'm going to do is solve this as a mini equation here. So we'll add nine on both sides of the equation, which will give me that x squared is equal to positive nine. Now what I'm going to do from here is take the square root on both sides of this equation, which will leave me with just x on the left side. Now what we're gonna have is that x cannot equal the positive and negative square root of nine, which is three. So x cannot equal three, but x can also not equal negative three. So this shows us exactly what our domain is going to look like. So our domain is going to go from negative infinity all the way up to negative three, not included because we're we can't have a negative three here. It's gonna go from negative three all the way up to positive three, not including that positive three. And then it's going to go from positive three all the way to positive infinity. So basically what I'm saying with this interval notation here is that my domain cannot be negative three and it cannot be positive three, but it can be any real number below negative three, above positive three, or anywhere between negative three and positive three, but it just can't be these specific values. So any way you choose to represent this domain, just make sure all of this information is included, and that is the solution for the domain of this problem, in this situation. So I hope you found this video helpful, and I'll see you all in the next one.

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)

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 $g (x)$ .
$g (x) = \frac{x + 1}{x^{2} - 9}$

$(- \infty, \infty)$

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

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

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

Verified step by step guidance

Identify the function given: $g(x) = \frac{x+1}{x^2 - 9}$. We need to find the domain, which consists of all real values of $x$ for which the function is defined.

Recall that a rational function is undefined where its denominator is zero. So, set the denominator equal to zero to find these values: $x^2 - 9 = 0$.

Solve the equation $x^2 - 9 = 0$ by factoring it as a difference of squares: $(x - 3)(x + 3) = 0$. This gives the solutions $x = 3$ and $x = -3$.

Since the function is undefined at $x = 3$ and $x = -3$, these values must be excluded from the domain.

Express the domain as all real numbers except $x = 3$ and $x = -3$, which in interval notation is $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$.

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