The domain of u(x) is the set of all real values except 0 and the domain of v(x) is the set of all real values except 2. What are the restrictions on the domain of (u circle v) (x)? u(x) Not-equals 0 and v(x) Not-equals 2
x Not-equals 0 and x cannot be any value for which u(x) Equals 2
x Not-equals 2 and x cannot be any value for which v(x) Equals 0
u(x) Not-equals 2 and v(x) Not-equals 0

The domain would include all the real values except and the x for which .

Hence, the restrictions would be:

and

Step-by-step explanation:

Given

The domain of u(x) is the set of all real values except 0.

so

domain = (-∞, 0) U (0, ∞)

The domain of v(x) is the set of all real values except 2.

so

domain = (-∞, 2) U (2, ∞)

For the function composed function (u circle v) (x), we need to apply first the function  whose argument is (x), and then the function whose argument is .

Please note that the domain of the composed function (u circle v) (x) must have to take into account the values for which both functions u(x) and v(x) are defined.

Therefore, the domain would include all the real values except and the x for which .

Hence, the restrictions would be:

and

24+x=y

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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