Find each of the following functions. f(x) = 5 − x , g(x) = x2 − 9 (a) f + g state the domain of the function. (enter your answer using interval notation.) (b) f − g state the domain of the function. (enter your answer using interval notation.) (c) fg incorrect: your answer is incorrect. state the domain of the function. (enter your answer using interval notation.) (d) f/g state the domain of the function. (enter your answer using interval notation.)

a, b and c= (-∞,∞) d = (-∞,3) U (-3,3) U (3,∞)

Step-by-step explanation:

Hi there!

1) Firstly, let's recap the sum of functions rule:

f(x) + g(x) = (f+g)(x)

Applying it to those functions, we have:

f(x)=5-x +g(x)= = [(5-x) +(-9)](x)

(f+g)(x)=

2) To State the Domain is to state the set which is valid the quantities of x, of a function. In this case,

a) (f+g)(x)= 5-x++9

Simplifiying

(f+g)(x) =

Since there are no restrictions neihter discontinuities, this function has a Domain which can expressed this way:

X may assume infinite quantities, negatives or positives one in the Real set.

(-∞< x <∞+) or simply (-∞,∞)

Or simply put, x ∈ R. Remember, ∞ is not a number, it's a notation meaning infinite values. That's why it's not a closed interval.

Check the graph below.

b) (f-g)(x) =(5-x) -(x²-9)

(f-g)(x)= 5-x-x²+9

Domain of (f-g)(x) =x²-x+14

Similarly to a) this function (f-g)(x) has not discontinuity, nor restrictions on its Domains.

Since there are no restrictions either discontinuities, this function has a Domain which can be expressed this way:

X may assume infinite quantities, negatives or positives one in the Real set.

(-∞< x <∞ +) then finally, the (-∞,∞)

c) (f*g)(x)=(5-x)(x²-9)

(f*g)(x)=5x²-45-x³+9x

Again, this function has no discontinuities, nor restrictions in its Domain as you can check it on its graph.

Then, the Domain of (f*g)(x)=(5-x)(x²-9) is also (-∞,∞)

d) (f/g)(x) =(5-x)/(x²-9)

Highlighting the denominator, we can calculate the Domain.

We can see a restriction here. There is no denominator zero, defined for the Set of R.

Then, let's calculate

and x < -3

In the Numerator, no restrictions.

So the Domain will be the union between the Numerator's Domain and the Denominator's Domain with Restrictions.

Check the graph below.

Finally

D = (-∞,3) U (-3,3) U (3,∞)

salary

step-by-step explanation:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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