Answer :
Answer:
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.
Step-by-step explanation:
From Function theory we get that range of a function is the set of values of [tex]y[/tex] so that exist a respective value of [tex]x[/tex], whose set is the domain of the function. Let be [tex]f(x) = (x+1)^{2}[/tex] and [tex]g(x) = -4\cdot |x+1|[/tex], 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:
[tex](x+1)^{2} \geq 0[/tex]
[tex](x+1)^{2}-4 \geq -4[/tex] (Compatibility with addition)
[tex]f(x) \geq -4[/tex] (Definition/Result)
The range of [tex]f(x)[/tex] is: [tex]Ran \{f(x)\} = [-4, +\infty)[/tex]
g(x):
From Real Algebra we know that a absolute value leads to non-negative number, which means that:
[tex]|x+1|\geq 0[/tex]
[tex]-4\cdot |x+1| \leq 0[/tex] ([tex]a> b \longrightarrow a \cdot c < b\cdot c[/tex], if [tex]c < 0[/tex])
[tex]g(x) \leq 0[/tex] (Definition/Result)
The range of [tex]g(x)[/tex] is: [tex]Ran\{g(x)\} = (-\infty, 0][/tex]
If we compare each range, we came into the conclusion that:
[tex]-4 \leq y \leq 0[/tex].
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.