Math  /  Algebra

Question[-/2 Points] DETAILS MY NOTES SCOLALG7 3.7.043.
Find the domain of the given function. (Enter your answer using interval notation.) h(x)=x414h(x)=\sqrt[4]{x^{4}-1} \square

Studdy Solution

STEP 1

What is this asking? We need to find all the *valid* inputs (xx) for the function h(x)=x414h(x) = \sqrt[4]{x^4 - 1}, meaning all the xx values that won't break the function! Watch out! We can't take the fourth root of a *negative* number!
So, we need to make sure that whatever is *inside* that radical is zero or positive!

STEP 2

1. Ensure the expression inside the radical is non-negative.
2. Solve the inequality.
3. Express the domain in interval notation.

STEP 3

Alright, so we know that whatever is *under* the fourth root has gotta be greater than or equal to zero.
Let's write that down mathematically: x410x^4 - 1 \ge 0 This makes sure we're only taking the fourth root of zero or a *positive* number, which is totally allowed!

STEP 4

We've got x410x^4 - 1 \ge 0.
This looks familiar, right?
It's a difference of squares!
Let's **factor** it: (x21)(x2+1)0(x^2 - 1)(x^2 + 1) \ge 0 Hey, look!
Another difference of squares!
Let's factor x21x^2 - 1: (x1)(x+1)(x2+1)0(x - 1)(x + 1)(x^2 + 1) \ge 0 Awesome!

STEP 5

Now, let's think about those factors.
Notice that x2+1x^2 + 1 is *always* positive, no matter what xx is!
Squaring a number always gives a positive (or zero), and then we add one.
So, that factor doesn't change the sign of the whole expression.
We can effectively *divide* both sides of the inequality by (x2+1)(x^2+1) to get 11, simplifying our inequality without changing the solution set: (x1)(x+1)0(x - 1)(x + 1) \ge 0

STEP 6

The inequality (x1)(x+1)0(x - 1)(x + 1) \ge 0 is equal to zero when x=1x = 1 or x=1x = -1.
These are our **critical points**.

STEP 7

Let's test some values around our critical points to see when the inequality is true. * If x<1x < -1 (let's try x=2x = -2), we get (21)(2+1)=(3)(1)=3>0(-2 - 1)(-2 + 1) = (-3)(-1) = 3 > 0.
It works! * If 1<x<1-1 < x < 1 (let's try x=0x = 0), we get (01)(0+1)=(1)(1)=1<0(0 - 1)(0 + 1) = (-1)(1) = -1 < 0.
Nope! * If x>1x > 1 (let's try x=2x = 2), we get (21)(2+1)=(1)(3)=3>0(2 - 1)(2 + 1) = (1)(3) = 3 > 0.
It works!

STEP 8

So, our inequality is true when x1x \le -1 or x1x \ge 1.
In interval notation, that's (,1][1,)(-\infty, -1] \cup [1, \infty).
Boom!

STEP 9

The domain of h(x)h(x) is (,1][1,)(-\infty, -1] \cup [1, \infty).

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