Question

If u (x) = negative 2 x squared and v (x) = StartFraction 1 Over x EndFraction, what is the range of (u circle v) (x)?

Answer 1

The range of (u circle v)(x) is all real numbers except 0.

Step-by-step explanation:

The range of (u circle v)(x) can be found by substituting the expression u(x) into the function v(x). The composition (u circle v)(x) is defined as u(v(x)), so we replace x in u(x) with v(x).

Given u(x) = -2x^2 and v(x) = 1/x, we have (u circle v)(x) = u(v(x)) = u(1/x) = -2(1/x)^2 = -2/x^2.

Therefore, the range of (u circle v)(x) is all real numbers except 0, since the function is undefined at x = 0.

Answer 2

(-∞, 0)

Step-by-step explanation:

Given

u(x) = -2x^2

v(x) = 1/x

Find

The range of (u◦v)(x)

Solution

(u◦v)(x) = u(v(x)) = u(1/x) = -2(-1/x)^2 = -2/x^2

The denominator is always positive (since x=0 is excluded from the domain), so the fraction is always negative. There is a horizontal asymptote at y=0. The range is ...

(-∞, 0)