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Math

Math Snap

PROBLEM

For f(x)=x2f(x) = x^2 and g(x)=x2+4g(x) = x^2 + 4, find the following composite functions and state the domain of each.
(a) fgf \circ g
(b) gfg \circ f
(c) fff \circ f
(d) ggg \circ g

STEP 1

What is this asking?
We need to find four composite functions, which basically means plugging one function into another, like a Russian nesting doll, and then figure out what values of xx are allowed.
Watch out!
Don't mix up the order of the functions when composing them! fgf \circ g means f(g(x))f(g(x)), not g(f(x))g(f(x)).
Also, be careful when finding the domain; think about what values of xx would cause any problems, like dividing by zero or taking the square root of a negative number.

STEP 2

1. Find fgf \circ g and its domain.
2. Find gfg \circ f and its domain.
3. Find fff \circ f and its domain.
4. Find ggg \circ g and its domain.

STEP 3

Alright, let's start with fgf \circ g, which means f(g(x))f(g(x)).
This is like putting the g(x)g(x) function inside the f(x)f(x) function!

STEP 4

We know f(x)=x2f(x) = x^2 and g(x)=x2+4g(x) = x^2 + 4.
So, f(g(x))f(g(x)) means we replace the xx in f(x)f(x) with the entire g(x)g(x) function.

STEP 5

So, f(g(x))=(g(x))2=(x2+4)2f(g(x)) = (g(x))^2 = (x^2 + 4)^2.
Let's expand this: (x2+4)2=x4+8x2+16(x^2 + 4)^2 = x^4 + 8x^2 + 16.

STEP 6

Now, for the domain.
Since we're only dealing with polynomials, there are no restrictions on xx. xx can be anything!
So, the domain of fgf \circ g is all real numbers, which we can write as (,)(-\infty, \infty).

STEP 7

Next up: gfg \circ f, which means g(f(x))g(f(x)).
This time, we're putting f(x)f(x) inside g(x)g(x).

STEP 8

We have g(x)=x2+4g(x) = x^2 + 4, so g(f(x))=(f(x))2+4g(f(x)) = (f(x))^2 + 4.
Since f(x)=x2f(x) = x^2, we get g(f(x))=(x2)2+4=x4+4g(f(x)) = (x^2)^2 + 4 = x^4 + 4.

STEP 9

Again, we have a polynomial, so the domain is all real numbers: (,)(-\infty, \infty).
No restrictions here!

STEP 10

Now for fff \circ f, meaning f(f(x))f(f(x)).
We're putting f(x)f(x) inside itself!
It's like function inception!

STEP 11

We have f(x)=x2f(x) = x^2, so f(f(x))=(f(x))2=(x2)2=x4f(f(x)) = (f(x))^2 = (x^2)^2 = x^4.

STEP 12

And, as you might expect, the domain is still all real numbers: (,)(-\infty, \infty).

STEP 13

Finally, let's tackle ggg \circ g, which is g(g(x))g(g(x)).
Putting g(x)g(x) inside itself!

STEP 14

We have g(x)=x2+4g(x) = x^2 + 4, so g(g(x))=(g(x))2+4=(x2+4)2+4g(g(x)) = (g(x))^2 + 4 = (x^2 + 4)^2 + 4.

STEP 15

Expanding this, we get (x2+4)2+4=x4+8x2+16+4=x4+8x2+20(x^2 + 4)^2 + 4 = x^4 + 8x^2 + 16 + 4 = x^4 + 8x^2 + 20.

STEP 16

Once again, a polynomial!
So, the domain is all real numbers: (,)(-\infty, \infty).

SOLUTION

(a) fg=x4+8x2+16f \circ g = x^4 + 8x^2 + 16, Domain: (,)(-\infty, \infty)
(b) gf=x4+4g \circ f = x^4 + 4, Domain: (,)(-\infty, \infty)
(c) ff=x4f \circ f = x^4, Domain: (,)(-\infty, \infty)
(d) gg=x4+8x2+20g \circ g = x^4 + 8x^2 + 20, Domain: (,)(-\infty, \infty)

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