Math  /  Algebra

Question5. Find the domain of the function: f(x)=x6(x+4)(x8)f(x)=\frac{x-6}{(x+4)(x-8)} (4,)(-4, \infty) (4,8)(8,(-4,8) \cup(-8, \propto (,4)(4(-\infty,-4) \cup(-4 (,4)(4(-\infty,-4) \cup(-4,
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Studdy Solution

STEP 1

What is this asking? We need to find all the *valid* inputs (xx values) for the function f(x)f(x), which is a fancy way of saying "where is this function defined?". Watch out! Remember, we can't divide by **zero**, so we need to be extra careful with the bottom of the fraction!

STEP 2

1. Find the trouble spots
2. Build the domain

STEP 3

Alright, let's **investigate** this function!
We've got f(x)=x6(x+4)(x8)f(x) = \frac{x-6}{(x+4)(x-8)}.
Now, fractions are usually cool, but they get a little *dramatic* when the denominator is zero.
So, our mission is to find any xx values that would cause this *denominator disaster*.

STEP 4

The denominator is (x+4)(x8)(x+4)(x-8).
We need to figure out when (x+4)(x8)=0(x+4)(x-8) = 0.
This happens when either x+4=0x+4=0 or x8=0x-8=0.

STEP 5

Let's tackle x+4=0x+4=0 first.
To isolate xx, we can add 4-4 to both sides of the equation.
This gives us x+4+(4)=0+(4)x + 4 + (-4) = 0 + (-4), which simplifies to x=4x = -4.
So, x=4x = \mathbf{-4} is one of our trouble spots!

STEP 6

Now, let's look at x8=0x-8=0.
Adding 88 to both sides gives us x8+8=0+8x - 8 + 8 = 0 + 8, which simplifies to x=8x = \mathbf{8}.
Another trouble spot found!

STEP 7

We've found our two trouble spots: x=4x=-4 and x=8x=8.
This means f(x)f(x) is defined for *every other* number!

STEP 8

So, the **domain** is all real numbers *except* 4-4 and 88.
We can write this using interval notation as (,4)(4,8)(8,)(-\infty, -4) \cup (-4, 8) \cup (8, \infty).
This basically says "all numbers less than 4-4, all numbers between 4-4 and 88, and all numbers greater than 88".
Perfect!

STEP 9

The domain of f(x)f(x) is (,4)(4,8)(8,)(-\infty, -4) \cup (-4, 8) \cup (8, \infty).

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