Given functions f and g, find the following :`(f+g)-f/g` . Also, find the domain.`f(x)=2x^2+3` ; `g(x)=4x^8+1`

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txmedteach | High School Teacher | (Level 3) Associate Educator

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Let's start off by looking at our function composition:

`F(x) = (f+g)-(f/g)`

We are also given:

`f(x) = 2x^2 + 3`

`g(x) = 4x^8+1`

Well, it looks to me like all we're going to do is substitute our functions into the function composition! Here's what that will look like:

`F(x) = (f+g)-(f/g) =((2x^2+3)+(4x^8+1)) - (2x^2+3)/(4x^8+1)`

We can actually simplify what's inside the parentheses:

`F(x) = 4x^8 + 2x^2 + 4 - (2x^2+3)/(4x^8+1)`

We might be done. However, often fr rational functions, you want to get everything over a single denominator. Let's go ahead and do that. We'll start by multiplying `4x^8+2x^2 + 4` by `(4x^8+1)/(4x^8+1)` in order to get a common denominator. Remember, we can't add without a common denominator! Keep in mind, too, when we subtract `(2x^2 + 3)`, we'll need to distribute that negative. All this gives us the following result:

`F(x) = ((4x^8+2x^2+4)(4x^8+1) - 2x^2-3)/(4x^8+1)`

Now, we'll simplify the terms above by multiplying the polynomials:

`F(x) = (16x^16+4x^8+8x^10+2x^2+16x^8+4 - 2x^2-3)/(4x^8+1)`

Unfortunately, only the `2x^2` terms cancel in the numerator, giving us a final function:

`F(x) = (16x^16 + 8x^10 + 20x^8 + 1)/(4x^8+1)`

I'm glad that's over.

Now to find the domain. Remember, to find the domain, we simply exclude the values for which `F(x)` is not defined. The only case in which `F(x)` is undefined is where the denominator is zero. So, let's solve for the x's that make the denominator zero:

`0 = 4x^8 + 1`

Well, I can tell already that if `x inRR` (x is a real number), then we're good. The domain is all real numbers because `4x^8+1` has no real roots. We can show this graphically below:

Notice that that blue line (`y = 4x^8+1)` never reaches zero. This means that there are no real roots. So, again, if we're only worried about the real numbers, the domain of F(x) is all real numbers (`x in RR`).

Now, there are 8 complex roots, which we can see here. I will not evaluate the roots of `i` because of the space requirement for answers, but they are easy to evaluate (I'll start you off, if you really want to do that).  Let's go back to setting our denominator to 0:

`4x^8+1 = 0`

We can factor the following two types of expressions with imaginary numbers:

`a^2 + b^2 = (a+bi)(a-bi)`

`a^2 - b^2 = (a+b)(a-b)`

Using these two we can factor our denominator into 8 factors. The process can be seen here:

`4x^8 + 1`


`= (sqrt2x^2 +isqrti)(sqrt2x^2-isqrti)(sqrt2x^2+sqrti)(sqrt2x^2-sqrti)`

Now, the next step is too long, so let me show the factors for each of the above terms individually so we can find the roots:

`sqrt2x^2+isqrti = (root(4)(2)x + isqrt(isqrti))(root(4)(2)x - isqrt(isqrti))`

`sqrt2x^2-isqrti = (root(4)(2)x +sqrt(isqrti))(root(4)(2)x-sqrt(isqrti))`

`sqrt2x^2+sqrti=(root(4)(2)x + iroot(4)(i))(root(4)(2)x - iroot(4)(i))`

`sqrt2x^2 - sqrti = (root(4)(2)x + root(4)(i))(root(4)(2)x - root(4)(i))`

Now, each term gives us a root for the denominator. The roots are listed below:

`x = +-root(4)(i/2), +- iroot(4)(i/2), +- sqrtiroot(4)(i/2), +-isqrtiroot(4)(i/2)`

So, our domain will become:

`{x in CC | x !in {+-root(4)(i/2), +- iroot(4)(i/2), +- sqrtiroot(4)(i/2), +-isqrtiroot(4)(i/2)}}`

I hope that helps! By the way, totally ignore all of those complex terms if you're only worried about the real domain.

Side Note: to calculate the roots of `i`, Just remember that `i = e^(pi/2*i).`

You can simply multiply exponents to find the desired number!