Question

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 the composite function (u \u2218 v)(x), given u(x) = -2x^2 and v(x) = 1/x, is all real numbers greater than 0.

Step-by-step explanation:

The question is asking for the range of the composite function (u \u2218 v)(x), where u(x) = -2x2 and v(x) = 1/x. To find the composite function, we substitute v(x) into u(x), getting u(v(x)) = u(1/x) = -2(1/x)2 = -2/x2. The range of -2/x2 is all real numbers greater than 0, because the function will never reach zero or negative values as x gets increasingly larger or smaller.

Answer 2

(−∞, 0) ∪ (0, ∞)

Step-by-step explanation:

Given :

u(x) = -2x²

`V(x)=\sqrt{(1)/(x) }`

Determining UoV(x) :

`UoV(x)=U(V(x))=-2\left( \sqrt{(1)/(x) } \right)^(2)`

Then

`UoV(x)=U(V(x))=-2 * (1)/(x)`

Then

`UoV(x)=U(V(x))=- (2)/(x)`

Since the range of -2/x is all real numbers except zero

Then

The range of UoV is :

(−∞, 0) ∪ (0, ∞)