Respuesta :

highwireact
Let's examine what the conditions on the domains of f and g have to be before we examine h. Since f is defined:

[tex]f(x)= \sqrt{2-x} [/tex], we know that the function will be undefined when 2 - x < 0, or when x > 2, so we have the restriction on our domain that x ≥ 0.

g(x) has no such restrictions and is defined for all possible values of x, so its domain is all real numbers, or [tex]\mathbb{R}[/tex]. 

Since h(x) is a rational function, we know that it's undefined when its denominator is 0. In this case, our denominator is g(x), so the function is undefined when g(x)=0. x - 1 = 0 only when x = 1, so the second restriction on our domain for h is that x ≠ 1. Putting those two together, we find that the domain of h(x) is all values of x ≥ 0, where x ≠ 1.
isyllus

Answer:

Domain of h(x) : [2,∞)

Step-by-step explanation:

Given: Two functions

[tex]f(x)=\sqrt{2-x}[/tex]

[tex]g(x)=x-1[/tex]

Composite function, [tex]h(x)=\dfrac{f(x)}{g(x)}[/tex]

[tex]f(x)=\sqrt{2-x}[/tex], It is square root function. As we know square root function is always greater than 0.

2-x ≥ 0

  2 ≥ x

Domain of f(x) : [2,∞)

[tex]g(x)=x-1[/tex], It is straight line equation (Linear function) As we know domain of linear function is all real number.

Domain of g(x) : (-∞,∞)

[tex]h(x)=\dfrac{f(x)}{g(x)}[/tex]

[tex]h(x)=\dfrac{\sqrt{2-x}}{x-1}[/tex] , It is rational function. Denominator can't be zero.

So, x-1 ≠ 0

        x ≠ 1

Now, we will see domain of h(x) common all three domain.

Domain of h(x): [2,∞)

Hence, The domain of h(x) is [2,∞)

Otras preguntas