Math / Algebra

QuestionConsider the following functions. $f(x) = \frac{1}{x}$ and $g(x) = x - 1$
Step 2 of 2: Find the formula for $(g \circ f)(x)$ and simplify your answer. Then find the domain for $(g \circ f)(x)$ . Round your answer to two decimal places, if necessary.
Answer 2 Points
$(g \circ f)(x) =$
Domain $=$

Studdy Solution

STEP 1

What is this asking? We're taking two functions, $f(x)$ and $g(x)$ , and we need to find the function $(g \circ f)(x)$ , which is the same as $g(f(x))$ , and then figure out what values of $x$ are allowed. Watch out! Remember, $(g \circ f)(x)$ means we're putting $f(x)$ inside $g(x)$ !
Also, be super careful about the domain; we need to make sure the resulting function makes sense!

STEP 2

1. Compose the functions
2. Find the domain

STEP 3

Alright, let's start by remembering what $(g \circ f)(x)$ actually means.
It means we're taking $f(x)$ and plugging it into $g(x)$ wherever we see an $x$ .
So, since $g(x) = x - 1$ , we're going to replace that $x$ with the whole function $f(x)$ , which is $\frac{1}{x}$ .

STEP 4

So, $(g \circ f)(x)$ becomes $g(f(x)) = \frac{1}{x} - 1$ .
Boom! That's our composed function!

STEP 5

Now, for the domain.
Remember, the domain is all the possible $x$ values we can plug in.
Let's look at our original functions.
For $f(x) = \frac{1}{x}$ , we can't have $x = 0$ because we can't divide by zero.
For $g(x) = x - 1$ , we can plug in anything!

STEP 6

But now, we have $(g \circ f)(x) = \frac{1}{x} - 1$ .
We still can't have $x = 0$ because of that $\frac{1}{x}$ .
So, our domain is all real numbers except $0$ !
We can write that as $(-\infty, 0) \cup (0, \infty)$ .
That means everything from negative infinity up to 0, and everything from 0 to positive infinity!

STEP 7

$(g \circ f)(x) = \frac{1}{x} - 1$
Domain $= (-\infty, 0) \cup (0, \infty)$

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Studdy Solution

STEP 1

STEP 2

1. Compose the functions
2. Find the domain

STEP 3

Alright, let's start by remembering what $(g \circ f)(x)$ actually means.
It means we're taking $f(x)$ and plugging it into $g(x)$ wherever we see an $x$ .
So, since $g(x) = x - 1$ , we're going to replace that $x$ with the whole function $f(x)$ , which is $\frac{1}{x}$ .

STEP 4

So, $(g \circ f)(x)$ becomes $g(f(x)) = \frac{1}{x} - 1$ .
Boom! That's our composed function!

STEP 5

Now, for the domain.
Remember, the domain is all the possible $x$ values we can plug in.
Let's look at our original functions.
For $f(x) = \frac{1}{x}$ , we can't have $x = 0$ because we can't divide by zero.
For $g(x) = x - 1$ , we can plug in anything!

STEP 6

STEP 7

$(g \circ f)(x) = \frac{1}{x} - 1$
Domain $= (-\infty, 0) \cup (0, \infty)$

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Studdy Solution

STEP 1

STEP 2

1. Compose the functions2. Find the domain

STEP 3

STEP 4

So, (g∘f)(x)(g \circ f)(x)(g∘f)(x) becomes g(f(x))=1x−1g(f(x)) = \frac{1}{x} - 1g(f(x))=x1​−1.Boom! That's our **composed function**!

STEP 5

Now, for the **domain**.Remember, the domain is all the possible xxx values we can plug in.Let's look at our original functions.For f(x)=1xf(x) = \frac{1}{x}f(x)=x1​, we can't have x=0x = 0x=0 because we can't divide by **zero**.For g(x)=x−1g(x) = x - 1g(x)=x−1, we can plug in *anything*!

STEP 6

STEP 7

(g∘f)(x)=1x−1(g \circ f)(x) = \frac{1}{x} - 1(g∘f)(x)=x1​−1Domain =(−∞,0)∪(0,∞)= (-\infty, 0) \cup (0, \infty)=(−∞,0)∪(0,∞)

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1. Compose the functions
2. Find the domain

So, $(g \circ f)(x)$ becomes $g(f(x)) = \frac{1}{x} - 1$ .
Boom! That's our composed function!

Now, for the domain.
Remember, the domain is all the possible $x$ values we can plug in.
Let's look at our original functions.
For $f(x) = \frac{1}{x}$ , we can't have $x = 0$ because we can't divide by zero.
For $g(x) = x - 1$ , we can plug in anything!

$(g \circ f)(x) = \frac{1}{x} - 1$
Domain $= (-\infty, 0) \cup (0, \infty)$