If u(x) =-2x^2 and v(x)=1 over x what is the range of (uov) (x)

u(x) = -2x², v(x)= 1/x

(u o v)(x) = u(v(x)) = -2(1/x)²= -2/x²

We can see that domain for x is going to bee all real numbers except 0. From the equation above we can see that graph of the function (u o v)(x) = -2/x² has horizontal asymptote y=0, because degree of numerator is less than degree of denominator ( (u o v)(x) = -2x⁰/x² ).

x² is always going to be positive, so range is going to be all negative numbers.

y<0, or y∈(0,-∞)

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !