1) F(x)= (X+2)/(X^2-4x-5)
A) Find the domain (-00,-1)u(-1,5)u(5,00)
B) find F(-3) I'm not sure how to do it.
C) Find F(X+2) I'm not sure either.
2. f(x)= (2x)/(x-4) G(x)= (x)/(X+5) find the domain of each.
A)(F+G) (4x^2+5x)(x-4)(X+5) x can't= 4,-5
B) (F*G) (2x^2)/(x-4)(x+5) domain is the same
C) (F/g) (2x^2)+5/x(x-4) Not sure how to interpret domain.
1B
Just plug in -3 for x.
F(-3) = (-3+2)/((-3)^2-4(-3)-5) = -1/16
1C
F(x+2) = ((x+2)+2)/((x+2)^2-4(x+2)-5) = (x+4)/(x^2-9)
2A
I get 3x(x+2) / (x-4)(x+5)
your domain is correct, however
2B ok
2C
f/g = 2(x+5) / (x-4)
So, it looks like the domain is all reals except x=4, but you have to remember that
g(x) is not defined for x = -5, so f/g is not defined.
g(x)=0 for x=0, so f/g is not defined there either
So, the domain is all reals except -5,0,4
1) F(x) = (x + 2) / (x^2 - 4x - 5)
A) To find the domain of a rational function, we need to consider any values of x that would make the denominator zero, since division by zero is undefined. In this case, the denominator is x^2 - 4x - 5.
To find the domain, we set the denominator equal to zero and solve for x:
x^2 - 4x - 5 = 0
We can factor this quadratic equation as:
(x - 5)(x + 1) = 0
Setting each factor equal to zero gives us:
x - 5 = 0 or x + 1 = 0
Solving these equations, we find:
x = 5 or x = -1
Therefore, the domain of F(x) is (-∞, -1) U (-1, 5) U (5, ∞).
B) To find F(-3), we substitute x = -3 into the given function:
F(-3) = (-3 + 2) / (-3^2 - 4(-3) - 5)
Simplifying inside the parentheses:
F(-3) = -1 / (9 + 12 - 5)
F(-3) = -1 / 16
So, F(-3) = -1/16.
C) To find F(x + 2), we substitute x + 2 into the given function:
F(x + 2) = ((x + 2) + 2) / ((x + 2)^2 - 4(x + 2) - 5)
Simplifying inside the parentheses:
F(x + 2) = (x + 4) / (x^2 + 4x + 4 - 4x - 8 - 5)
F(x + 2) = (x + 4) / (x^2 - 9)
Therefore, F(x + 2) = (x + 4)/(x^2 - 9).
2) f(x) = 2x/(x - 4), g(x) = x/(x + 5)
A) To find the domain of each function, we need to consider any values of x that would make the denominator zero, since division by zero is undefined.
For f(x), the denominator is x - 4. Therefore, x cannot be equal to 4 since it would make the denominator zero.
For g(x), the denominator is x + 5. Therefore, x cannot be equal to -5 since it would make the denominator zero.
So, the domain of f(x) is all real numbers except x = 4, and the domain of g(x) is all real numbers except x = -5.
B) (f + g)(x) = (2x/(x - 4)) + (x/(x + 5))
To find the domain of this sum, we need to consider the domain restrictions from f(x) and g(x). In this case, the domain would be the intersection of the domains of f(x) and g(x), which means x cannot be equal to 4 or -5.
Therefore, the domain of (f + g)(x) is all real numbers except x = 4 and x = -5.
C) (f * g)(x) = (2x/(x - 4)) * (x/(x + 5))
To find the domain of this product, we consider the domain restrictions from f(x) and g(x). In this case, the domain would be the intersection of the domains of f(x) and g(x), which means x cannot be equal to 4 or -5.
Therefore, the domain of (f * g)(x) is all real numbers except x = 4 and x = -5.
D) (f / g)(x) = (2x/(x - 4)) / (x/(x + 5))
To find the domain of this quotient, we consider the domain restrictions from f(x) and g(x). In this case, the domain would be the intersection of the domain of f(x) and the set of all x-values where g(x) is not equal to zero.
Since g(x) is nonzero everywhere except x = -5, we only need to exclude x = 4 from the domain.
Therefore, the domain of (f / g)(x) is all real numbers except x = 4.