Question

If y(x) = -2x^2 + 3 and v(x) = -√x, what is the range of (V°v)(x)?

a) (3.00)

b) (-0,3)

c) (-[infinity],0)

d) (-[infinity],00)

Answer 1

The range is (-∞, 0).

The correct option is c) (-∞, 0)

Step-by-step explanation:

The composition of functions
`\(V \circ v\)`is found by taking the function
`V(x)\) and replacing \(x\) with \(v(x)\)`. Mathematically, this is expressed as
`\((V \circ v)(x) = V(v(x))\).`

Given
`\(V(x) = -√(x)\), substituting \(v(x) = -2x^2 + 3\) into \(V(x)\) gives \((V \circ v)(x) = -√(-2x^2 + 3)\).`

To determine the range of
`\((V \circ v)(x)\),` we need to analyze the behavior of the inner function
`\(-2x^2 + 3\)` under the square root. The expression
`\(-2x^2 + 3\)`represents a downward-facing parabola, and the square root of this expression is defined for values where
`\(-2x^2 + 3 \geq 0\).`

Solving
`\(-2x^2 + 3 \geq 0\) gives \(x^2 \leq (3)/(2)\), which implies \(-\sqrt{(3)/(2)} \leq x \leq \sqrt{(3)/(2)}\).`

However, since we are looking for the range of
`\((V \circ v)(x) = -√(-2x^2 + 3)\),` the negative square root introduces an additional negation. Therefore, the range is
`\((-∞, 0)\)`. This means that for any valid input
`\(x\), \((V \circ v)(x)\)`will produce a value less than or equal to zero.

The correct option is c) (-∞, 0)