If u(x)=-2x^2+3 and (x)=1/x, what is the range of (u°v)(x)​

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.