Respuesta :
Answer:
(a) [tex](f+g)(x)=x^2-x-1[/tex]; Domain = (-∞,∞)
(b) [tex](f-g)(x)=-x^2-x+7[/tex]; Domain = (-∞,∞)
(c) [tex](fg)(x)=-x^3+3x^2+4x-12[/tex]; Domain = (-∞,∞)
(d) [tex](\frac{f}{g})(x)=\frac{3-x}{x^2-4}[/tex]; Domain = (-∞,-2)∪(-2,2)∪(2,∞)
Step-by-step explanation:
The given functions are
[tex]f(x)=3-x[/tex]
[tex]g(x)=x^2-4[/tex]
(a)
We need to find the function f+g.
[tex](f+g)(x)=f(x)+g(x)[/tex]
[tex](f+g)(x)=(3-x)+(x^2-4)[/tex]
[tex](f+g)(x)=x^2-x-1[/tex]
It is a polynomial and domain of a polynomial is all real numbers.
Domain = (-∞,∞)
(b)
We need to find the function f-g.
[tex](f-g)(x)=f(x)-g(x)[/tex]
[tex](f-g)(x)=(3-x)-(x^2-4)[/tex]
[tex](f-g)(x)=3-x-x^2+4[/tex]
[tex](f-g)(x)=-x^2-x+7[/tex]
It is a polynomial. So,
Domain = (-∞,∞)
(c)
We need to find the function fg.
[tex](fg)(x)=f(x)g(x)[/tex]
[tex](fg)(x)=(3-x)(x^2-4)[/tex]
[tex](fg)(x)=3(x^2-4)-x(x^2-4)[/tex]
[tex](fg)(x)=3x^2-12-x^3+4x[/tex]
[tex](fg)(x)=-x^3+3x^2+4x-12[/tex]
It is a polynomial. So,
Domain = (-∞,∞)
(d)
We need to find the function f/g.
[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}[/tex]
[tex](\frac{f}{g})(x)=\frac{3-x}{x^2-4}[/tex]
It is a rational function. Domain of a rational function is all real except those values at which the denominator is equal to 0.
Equate denominator equal to 0.
[tex]x^2-4=0[/tex]
[tex]x^2=4[/tex]
Taking square both sides.
[tex]x=\pm \sqrt{4}[/tex]
[tex]x=\pm 2[/tex]
The function is not defined for x=-2 and x=2. So, domain of f/g is all real numbers except -2 and 2.
Domain = (-∞,-2)∪(-2,2)∪(2,∞)
Answer with Step-by-step explanation:
We are given that [tex]f(x)=3x-2,g(x)=x^2-4[/tex]
We have to find the
a.Domain of f+g
Domain of f(x)=([tex]-\infty,\infty[/tex])
Suppose y=g(x)=[tex]x^2-4[/tex]
[tex]x=\sqrt{y+4}[/tex]
Therefore, [tex]g^{-1}(x)=\sqrt{x+4}[/tex]
Domain of [tex]g^{-1}(x)=[-4,\infty)[/tex]
Range of [tex]g^{-1}(x)=[0,\infty)[/tex]
Range of [tex]g^{-1}(x)[/tex]=Domain of g(x)=[tex][0,\infty)[/tex]
Domain of (f+g)(x)=[0,[tex]\infty[/tex])
Because domain of sum of two function is the intersection of domain of given functions.
Similar, domain of (f-g)(x)=[0,[tex]\infty[/tex])
Because domain of difference of two function is the intersection of domain of given functions.
Similarly , domain of fg(x)=[0,[tex]\infty[/tex])
Because domain of product of two function is the intersection of domain of given functions.
[tex]\frac{f}{g}(x)=\frac{3-x}{x^2-4}=\frac{3-x}{(x-2)(x+2)}[/tex]
Function is not define at x=2 and x=-2
Therefore, domain of [tex]\frac{f}{g}(x)[/tex]=R-{-2,2}=[tex](-\infty,-2)\cup(-2,2)\cup(2,\infty)[/tex]
