Question

f(x) = (x + 1)2 – 4 g(x) = –4 |x + 1| Which statement compares the ranges of the functions? The ranges of both f(x) and g(x) are limited to y ≥ –1. The ranges of both f(x) and g(x) are limited to y ≤ –1. Since the range of f(x) is limited to y ≤ 0 and the range of g(x) is limited to y ≥ –4, the ranges overlap for the values –4 ≤ y ≤ 0. Since the range of f(x) is limited to y ≥ –4 and the range of g(x) is limited to y ≤ 0, the ranges overlap for the values –4 ≤ y ≤ 0.

Answer 1

Its D

Explanation:

Got it right on EDGE

Answer 2

Since the range of f(x) is limited to y ≥ –4 and the range of g(x) is limited to y ≤ 0, the ranges overlap for the values –4 ≤ y ≤ 0.

Explanation:

From Function theory we get that range of a function is the set of values of
`y` so that exist a respective value of
`x`, whose set is the domain of the function. Let be
`f(x) = (x+1)^(2)` and
`g(x) = -4\cdot |x+1|`, the ranges of each function are, respectively:

f(x):

From Real Algebra we know that the square of every number is a non-negative number, which means that:

`(x+1)^(2) \geq 0`

`(x+1)^(2)-4 \geq -4` (Compatibility with addition)

`f(x) \geq -4` (Definition/Result)

The range of
`f(x)` is:
`Ran \{f(x)\} = [-4, +\infty)`

g(x):

From Real Algebra we know that a absolute value leads to non-negative number, which means that:

`|x+1|\geq 0`

`-4\cdot |x+1| \leq 0` (
`a> b \longrightarrow a \cdot c < b\cdot c`, if
`c < 0`)

`g(x) \leq 0` (Definition/Result)

The range of
`g(x)` is:
`Ran\{g(x)\} = (-\infty, 0]`

If we compare each range, we came into the conclusion that:

`-4 \leq y \leq 0`.

Thus, correct answer is: Since the range of f(x) is limited to y ≥ –4 and the range of g(x) is limited to y ≤ 0, the ranges overlap for the values –4 ≤ y ≤ 0.