Answer:
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)=[tex]x^{2} -9[/tex] = [(5-x) +([tex]x^{2}[/tex]-9)](x)
(f+g)(x)=[tex]5-x+x^{2} -9[/tex]
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+[tex]x^{2}[/tex]+9
Simplifiying
(f+g)(x) =[tex]x^{2} -x+4=0[/tex]
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 answer: (-∞,∞)
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
[tex]x^{2} -9>0\\ \sqrt{x^2} >\sqrt{9}\\x>3[/tex] 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,∞)
