If [tex]\( r(x) = 2 - x^2 \)[/tex] and [tex]\( w(x) = x - 2 \)[/tex], what is the range of [tex]\( (w \circ r)(x) \)[/tex]?

A. [tex]\((-\infty, 0]\)[/tex]
B. [tex]\((-\infty, 2]\)[/tex]
C. [tex]\([0, \infty)\)[/tex]
D. [tex]\([2, \infty)\)[/tex]



Answer :

To find the range of [tex]\((w \circ r)(x)\)[/tex], we need to determine the composition of the functions [tex]\(w(x)\)[/tex] and [tex]\(r(x)\)[/tex]. The composition [tex]\(w(r(x))\)[/tex] involves substituting [tex]\(r(x)\)[/tex] into [tex]\(w(x)\)[/tex].

Given:
[tex]\[ r(x) = 2 - x^2 \][/tex]
[tex]\[ w(x) = x - 2 \][/tex]

First, let's find [tex]\(w(r(x))\)[/tex]:
[tex]\[ w(r(x)) = w(2 - x^2) \][/tex]

Since [tex]\(w(x)\)[/tex] is defined as [tex]\(x - 2\)[/tex], substitute [tex]\(2 - x^2\)[/tex] into [tex]\(w\)[/tex]:

[tex]\[ w(2 - x^2) = (2 - x^2) - 2 \][/tex]
[tex]\[ w(2 - x^2) = 2 - x^2 - 2 \][/tex]
[tex]\[ w(2 - x^2) = -x^2 \][/tex]

Next, we need to determine the range of the function [tex]\(w(r(x)) = -x^2\)[/tex].

The quadratic function [tex]\(-x^2\)[/tex] opens downwards because the coefficient of [tex]\(x^2\)[/tex] is negative. For any real number [tex]\(x\)[/tex], [tex]\(x^2\)[/tex] is always non-negative (i.e., [tex]\(x^2 \geq 0\)[/tex]). Therefore, [tex]\(-x^2\)[/tex] will be non-positive and will achieve its maximum value at 0 when [tex]\(x = 0\)[/tex].

Thus, [tex]\(-x^2\)[/tex] for all real [tex]\(x\)[/tex] takes all values in the interval [tex]\((-\infty, 0]\)[/tex].

Therefore, the range of [tex]\((w \circ r)(x)\)[/tex] is:
[tex]\[ \boxed{(-\infty, 0]} \][/tex]