Math / Algebra

Questionnlist 0 Points: 0 of 1
For the given functions, find $(f \circ g)(x)$ and $(g \circ f)(x)$ and the domain of each. $f(x)=3 x+1, g(x)=\sqrt{x}$ 36 $(f \circ g)(x)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find two new functions: f(g(x)) and g(f(x)), and also figure out what values of x are allowed for each. Watch out! Square roots can be tricky with negative numbers, so we'll need to keep an eye on the domain!

STEP 2

1. Find f(g(x))
2. Find the domain of f(g(x))
3. Find g(f(x))
4. Find the domain of g(f(x))

STEP 3

Let's start by remembering what $(f \circ g)(x)$ actually means.
It means we're putting $g(x)$ inside $f(x)$ !
It's like a function turducken!

STEP 4

We know that $f(x) = 3x + 1$ and $g(x) = \sqrt{x}$ .
So, to find $f(g(x))$ , we replace the x in $f(x)$ with the entire $g(x)$ function.

STEP 5

This gives us $f(g(x)) = 3(\sqrt{x}) + 1$ .
See how we substituted $\sqrt{x}$ for x?

STEP 6

Now, the domain is all the possible x values we can plug in without breaking math.
We can't take the square root of a negative number, can we?

STEP 7

So, for $\sqrt{x}$ to be happy, x must be greater than or equal to zero.
That means our domain is $x \ge 0$ .

STEP 8

Time to flip the script and find $(g \circ f)(x)$ .
Now we're putting $f(x)$ inside $g(x)$ !

STEP 9

We substitute $f(x)$ which is $3x + 1$ into $g(x)$ which gives us $g(f(x)) = \sqrt{3x + 1}$ .

STEP 10

Again, we need to make sure the stuff inside the square root isn't negative.
So, we need $3x + 1 \ge 0$ .

STEP 11

Let's solve for x! Subtract 1 from both sides: $3x \ge -1$ .

STEP 12

Now, divide both sides by 3: $x \ge -\frac{1}{3}$ .
So, our domain is $x \ge -\frac{1}{3}$ .

STEP 13

$(f \circ g)(x) = 3\sqrt{x} + 1$ with domain $x \ge 0$
$(g \circ f)(x) = \sqrt{3x + 1}$ with domain $x \ge -\frac{1}{3}$

Was this helpful?

Questionnlist 0 Points: 0 of 1
For the given functions, find $(f \circ g)(x)$ and $(g \circ f)(x)$ and the domain of each. $f(x)=3 x+1, g(x)=\sqrt{x}$ 36 $(f \circ g)(x)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find two new functions: f(g(x)) and g(f(x)), and also figure out what values of x are allowed for each. Watch out! Square roots can be tricky with negative numbers, so we'll need to keep an eye on the domain!

STEP 2

1. Find f(g(x))
2. Find the domain of f(g(x))
3. Find g(f(x))
4. Find the domain of g(f(x))

STEP 3

Let's start by remembering what $(f \circ g)(x)$ actually means.
It means we're putting $g(x)$ inside $f(x)$ !
It's like a function turducken!

STEP 4

We know that $f(x) = 3x + 1$ and $g(x) = \sqrt{x}$ .
So, to find $f(g(x))$ , we replace the x in $f(x)$ with the entire $g(x)$ function.

STEP 5

This gives us $f(g(x)) = 3(\sqrt{x}) + 1$ .
See how we substituted $\sqrt{x}$ for x?

STEP 6

Now, the domain is all the possible x values we can plug in without breaking math.
We can't take the square root of a negative number, can we?

STEP 7

So, for $\sqrt{x}$ to be happy, x must be greater than or equal to zero.
That means our domain is $x \ge 0$ .

STEP 8

Time to flip the script and find $(g \circ f)(x)$ .
Now we're putting $f(x)$ inside $g(x)$ !

STEP 9

We substitute $f(x)$ which is $3x + 1$ into $g(x)$ which gives us $g(f(x)) = \sqrt{3x + 1}$ .

STEP 10

Again, we need to make sure the stuff inside the square root isn't negative.
So, we need $3x + 1 \ge 0$ .

STEP 11

Let's solve for x! Subtract 1 from both sides: $3x \ge -1$ .

STEP 12

Now, divide both sides by 3: $x \ge -\frac{1}{3}$ .
So, our domain is $x \ge -\frac{1}{3}$ .

STEP 13

$(f \circ g)(x) = 3\sqrt{x} + 1$ with domain $x \ge 0$
$(g \circ f)(x) = \sqrt{3x + 1}$ with domain $x \ge -\frac{1}{3}$

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Questionnlist 0 Points: 0 of 1For the given functions, find (f∘g)(x)(f \circ g)(x)(f∘g)(x) and (g∘f)(x)(g \circ f)(x)(g∘f)(x) and the domain of each. f(x)=3x+1,g(x)=xf(x)=3 x+1, g(x)=\sqrt{x}f(x)=3x+1,g(x)=x​ 36 (f∘g)(x)=(f \circ g)(x)=(f∘g)(x)= □\square□

Studdy Solution

STEP 1

What is this asking? We need to find two new functions: *f(g(x))* and *g(f(x))*, and also figure out what values of *x* are allowed for each. Watch out! Square roots can be tricky with negative numbers, so we'll need to keep an eye on the domain!

STEP 2

1. Find *f(g(x))*2. Find the domain of *f(g(x))*3. Find *g(f(x))*4. Find the domain of *g(f(x))*

STEP 3

Let's **start** by remembering what (f∘g)(x)(f \circ g)(x)(f∘g)(x) *actually means*.It means we're putting g(x)g(x)g(x) *inside* f(x)f(x)f(x)!It's like a function *turducken*!

STEP 4

We know that f(x)=3x+1f(x) = 3x + 1f(x)=3x+1 and g(x)=xg(x) = \sqrt{x}g(x)=x​.So, to find f(g(x))f(g(x))f(g(x)), we **replace** the *x* in f(x)f(x)f(x) with the *entire* g(x)g(x)g(x) function.

STEP 5

This gives us f(g(x))=3(x)+1f(g(x)) = 3(\sqrt{x}) + 1f(g(x))=3(x​)+1.See how we **substituted** x\sqrt{x}x​ for *x*?

STEP 6

Now, the **domain** is all the possible *x* values we can plug in *without breaking math*.We can't take the square root of a negative number, can we?

STEP 7

So, for x\sqrt{x}x​ to be happy, *x* must be greater than or equal to zero.That means our **domain** is x≥0x \ge 0x≥0.

STEP 8

Time to flip the script and find (g∘f)(x)(g \circ f)(x)(g∘f)(x).Now we're putting f(x)f(x)f(x) *inside* g(x)g(x)g(x)!

STEP 9

We **substitute** f(x)f(x)f(x) which is 3x+13x + 13x+1 into g(x)g(x)g(x) which gives us g(f(x))=3x+1g(f(x)) = \sqrt{3x + 1}g(f(x))=3x+1​.

STEP 10

Again, we need to make sure the stuff *inside* the square root isn't negative.So, we need 3x+1≥03x + 1 \ge 03x+1≥0.

STEP 11

Let's **solve** for *x*! **Subtract** 1 from both sides: 3x≥−13x \ge -13x≥−1.

STEP 12

Now, **divide** both sides by 3: x≥−13x \ge -\frac{1}{3}x≥−31​.So, our **domain** is x≥−13x \ge -\frac{1}{3}x≥−31​.

STEP 13

(f∘g)(x)=3x+1(f \circ g)(x) = 3\sqrt{x} + 1(f∘g)(x)=3x​+1 with domain x≥0x \ge 0x≥0(g∘f)(x)=3x+1(g \circ f)(x) = \sqrt{3x + 1}(g∘f)(x)=3x+1​ with domain x≥−13x \ge -\frac{1}{3}x≥−31​

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Questionnlist 0 Points: 0 of 1
For the given functions, find $(f \circ g)(x)$ and $(g \circ f)(x)$ and the domain of each. $f(x)=3 x+1, g(x)=\sqrt{x}$ 36 $(f \circ g)(x)=$ $\square$

What is this asking? We need to find two new functions: f(g(x)) and g(f(x)), and also figure out what values of x are allowed for each. Watch out! Square roots can be tricky with negative numbers, so we'll need to keep an eye on the domain!

1. Find f(g(x))
2. Find the domain of f(g(x))
3. Find g(f(x))
4. Find the domain of g(f(x))

Let's start by remembering what $(f \circ g)(x)$ actually means.
It means we're putting $g(x)$ inside $f(x)$ !
It's like a function turducken!

We know that $f(x) = 3x + 1$ and $g(x) = \sqrt{x}$ .
So, to find $f(g(x))$ , we replace the x in $f(x)$ with the entire $g(x)$ function.

This gives us $f(g(x)) = 3(\sqrt{x}) + 1$ .
See how we substituted $\sqrt{x}$ for x?

Now, the domain is all the possible x values we can plug in without breaking math.
We can't take the square root of a negative number, can we?

So, for $\sqrt{x}$ to be happy, x must be greater than or equal to zero.
That means our domain is $x \ge 0$ .

Time to flip the script and find $(g \circ f)(x)$ .
Now we're putting $f(x)$ inside $g(x)$ !

We substitute $f(x)$ which is $3x + 1$ into $g(x)$ which gives us $g(f(x)) = \sqrt{3x + 1}$ .

Again, we need to make sure the stuff inside the square root isn't negative.
So, we need $3x + 1 \ge 0$ .

Let's solve for x! Subtract 1 from both sides: $3x \ge -1$ .

Now, divide both sides by 3: $x \ge -\frac{1}{3}$ .
So, our domain is $x \ge -\frac{1}{3}$ .

$(f \circ g)(x) = 3\sqrt{x} + 1$ with domain $x \ge 0$
$(g \circ f)(x) = \sqrt{3x + 1}$ with domain $x \ge -\frac{1}{3}$