Respuesta :
Answer:
The range is all real number y<3.
Step-by-step explanation:
[tex](u \circ v)(x)=u(v(x))[/tex]
So we have to have v(x) exist for input x.
Let's think about that. v(x)=1/x so the domain is all real numbers except 0 since you cannot divide by 0. v(x)=1/x will also never output 0 because the numerator of 1/x is never 0. So the range of v(x)=1/x is also all real numbers except y=0.
Now let's plug v into u:
[tex]u(x)=-2x^2+3[/tex]
[tex]u(v(x))=u(\frac{1}{x}[/tex]
[tex]u(v(x))=-2(\frac{1}{x})^2+3[/tex]
The domain of will still have the restrictions of v; let's see if we see any others here.
Nope, there are no, others, the only thing that is bothering this function is still the division by x (which means we can't plug in 0).
[tex]u(v(x))=\frac{-2}{x^2}+3[/tex]
Let's thing about what are y's value will not ever get to be.
Let's start with that fraction. -2/x^2 will never be 0 because -2 will never be 0.
So we will never have y=0+3 which means y will never be 3.
There is one more thing to notice -2/x^2 will never be positive because x^2 is always positive and as we know a negative divided by a positive is negative.
So we have (a always negative number) + 3 this means the range will only go as high as 3 without including 3.
The range is all real number y<3.
