Math  /  Algebra

QuestionGiven that f(x)=x26xf(x)=x^{2}-6 x and g(x)=x+7g(x)=x+7, find and simplify the following: a. (f+g)(x)=(f+g)(x)= \square and the domain of (f+g)(f+g) is b. (fg)(x)=(f-g)(x)= \square and the domain of (fg)(f-g) is c. (fg)(x)=(f g)(x)= \square and the domain of (fg)(f g) is d. (fg)(x)=\left(\frac{f}{g}\right)(x)= \square and the domain of (fg)\left(\frac{f}{g}\right) is

Studdy Solution

STEP 1

What is this asking? We're given two functions, f(x)f(x) and g(x)g(x), and we need to find their sum, difference, product, and quotient, along with the domains of the resulting functions. Watch out! Remember that the domain of a function is the set of all possible xx values for which the function is defined.
Be careful when dividing by zero!

STEP 2

1. Find (f+g)(x)(f+g)(x) and its domain.
2. Find (fg)(x)(f-g)(x) and its domain.
3. Find (fg)(x)(fg)(x) and its domain.
4. Find (fg)(x)\left(\frac{f}{g}\right)(x) and its domain.

STEP 3

We are given f(x)=x26xf(x) = x^2 - 6x and g(x)=x+7g(x) = x + 7.

STEP 4

To find (f+g)(x)(f+g)(x), we simply add the two functions together: (f+g)(x)=f(x)+g(x)=(x26x)+(x+7)=x25x+7.(f+g)(x) = f(x) + g(x) = (x^2 - 6x) + (x + 7) = x^2 - 5x + 7.

STEP 5

Since (f+g)(x)(f+g)(x) is a polynomial, its domain is all real numbers.
We can write this as (,)(-\infty, \infty).

STEP 6

Now, let's find (fg)(x)(f-g)(x) by subtracting g(x)g(x) from f(x)f(x): (fg)(x)=f(x)g(x)=(x26x)(x+7)=x27x7.(f-g)(x) = f(x) - g(x) = (x^2 - 6x) - (x + 7) = x^2 - 7x - 7.

STEP 7

Again, since (fg)(x)(f-g)(x) is a polynomial, its domain is all real numbers, or (,)(-\infty, \infty).

STEP 8

To find (fg)(x)(fg)(x), we multiply the two functions: (fg)(x)=f(x)g(x)=(x26x)(x+7)=x3+7x26x242x=x3+x242x.(fg)(x) = f(x) \cdot g(x) = (x^2 - 6x)(x + 7) = x^3 + 7x^2 - 6x^2 - 42x = x^3 + x^2 - 42x.

STEP 9

As (fg)(x)(fg)(x) is also a polynomial, its domain is all real numbers, (,)(-\infty, \infty).

STEP 10

Finally, let's find (fg)(x)\left(\frac{f}{g}\right)(x) by dividing f(x)f(x) by g(x)g(x): (fg)(x)=f(x)g(x)=x26xx+7.\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2 - 6x}{x + 7}.

STEP 11

Here, we need to be careful!
The denominator cannot be zero.
So, we must exclude any xx value that makes g(x)=0g(x) = 0.
We have x+7=0x + 7 = 0, which means x=7x = -7.
Therefore, the domain of (fg)(x)\left(\frac{f}{g}\right)(x) is all real numbers *except* x=7x = -7.
We can write this as (,7)(7,)(-\infty, -7) \cup (-7, \infty).

STEP 12

a. (f+g)(x)=x25x+7(f+g)(x) = x^2 - 5x + 7, Domain: (,)(-\infty, \infty) b. (fg)(x)=x27x7(f-g)(x) = x^2 - 7x - 7, Domain: (,)(-\infty, \infty) c. (fg)(x)=x3+x242x(fg)(x) = x^3 + x^2 - 42x, Domain: (,)(-\infty, \infty) d. (fg)(x)=x26xx+7\left(\frac{f}{g}\right)(x) = \frac{x^2 - 6x}{x + 7}, Domain: (,7)(7,)(-\infty, -7) \cup (-7, \infty)

Was this helpful?

Studdy solves anything!

banner

Start learning now

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

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord