Answer :
To determine the range of [tex]\((u \circ v)(x)\)[/tex], we first need to find the composition of the two functions [tex]\(u(x)\)[/tex] and [tex]\(v(x)\)[/tex].
Given:
[tex]\[ u(x) = -2x^2 + 3 \][/tex]
[tex]\[ v(x) = \frac{1}{x} \][/tex]
The composition [tex]\((u \circ v)(x)\)[/tex] means we substitute [tex]\(v(x)\)[/tex] into [tex]\(u(x)\)[/tex]:
[tex]\[ (u \circ v)(x) = u(v(x)) = u\left(\frac{1}{x}\right) \][/tex]
So, we substitute [tex]\(\frac{1}{x}\)[/tex] into [tex]\(u(x)\)[/tex]:
[tex]\[ u\left(\frac{1}{x}\right) = -2\left(\frac{1}{x}\right)^2 + 3 \][/tex]
[tex]\[ = -2\left(\frac{1}{x^2}\right) + 3 \][/tex]
[tex]\[ = -\frac{2}{x^2} + 3 \][/tex]
Now we need to determine the range of the function [tex]\( -\frac{2}{x^2} + 3 \)[/tex]. To do this, we analyze the expression:
1. Analyze [tex]\(-\frac{2}{x^2}\)[/tex]:
- The function [tex]\(x^2\)[/tex] is always positive except when [tex]\(x = 0\)[/tex], where it is undefined.
- Thus, [tex]\(\frac{1}{x^2}\)[/tex] is also always positive for all [tex]\(x \neq 0\)[/tex].
- Since [tex]\(\frac{1}{x^2} > 0\)[/tex], [tex]\(-\frac{2}{x^2} < 0\)[/tex].
2. Analyze the whole expression [tex]\(-\frac{2}{x^2} + 3\)[/tex]:
- Because [tex]\(\frac{2}{x^2}\)[/tex] is always positive, [tex]\(-\frac{2}{x^2}\)[/tex] is always negative.
- Therefore, [tex]\(-\frac{2}{x^2} + 3\)[/tex] will always be less than 3 (since we are subtracting a positive quantity from 3).
- As [tex]\(x\)[/tex] approaches [tex]\(\pm\infty\)[/tex], [tex]\(\frac{1}{x^2}\)[/tex] approaches 0 and [tex]\(-\frac{2}{x^2}\)[/tex] approaches 0. Therefore, [tex]\(-\frac{2}{x^2} + 3\)[/tex] will approach 3 but never actually reach 3.
- As [tex]\(x\)[/tex] approaches 0 (from either the positive or negative direction), [tex]\(\frac{1}{x^2}\)[/tex] approaches [tex]\(\infty\)[/tex] and [tex]\(-\frac{2}{x^2}\)[/tex] approaches [tex]\(-\infty\)[/tex], making [tex]\(-\frac{2}{x^2} + 3\)[/tex] become very negative.
Hence, the value of [tex]\(-\frac{2}{x^2} + 3\)[/tex] can be any real number less than 3.
Therefore, the range of the composite function [tex]\((u \circ v)(x)\)[/tex] is:
[tex]\[ (-\infty, 3) \][/tex]
Thus, the correct answer is:
[tex]\[ (-\infty, 3) \][/tex]
Given:
[tex]\[ u(x) = -2x^2 + 3 \][/tex]
[tex]\[ v(x) = \frac{1}{x} \][/tex]
The composition [tex]\((u \circ v)(x)\)[/tex] means we substitute [tex]\(v(x)\)[/tex] into [tex]\(u(x)\)[/tex]:
[tex]\[ (u \circ v)(x) = u(v(x)) = u\left(\frac{1}{x}\right) \][/tex]
So, we substitute [tex]\(\frac{1}{x}\)[/tex] into [tex]\(u(x)\)[/tex]:
[tex]\[ u\left(\frac{1}{x}\right) = -2\left(\frac{1}{x}\right)^2 + 3 \][/tex]
[tex]\[ = -2\left(\frac{1}{x^2}\right) + 3 \][/tex]
[tex]\[ = -\frac{2}{x^2} + 3 \][/tex]
Now we need to determine the range of the function [tex]\( -\frac{2}{x^2} + 3 \)[/tex]. To do this, we analyze the expression:
1. Analyze [tex]\(-\frac{2}{x^2}\)[/tex]:
- The function [tex]\(x^2\)[/tex] is always positive except when [tex]\(x = 0\)[/tex], where it is undefined.
- Thus, [tex]\(\frac{1}{x^2}\)[/tex] is also always positive for all [tex]\(x \neq 0\)[/tex].
- Since [tex]\(\frac{1}{x^2} > 0\)[/tex], [tex]\(-\frac{2}{x^2} < 0\)[/tex].
2. Analyze the whole expression [tex]\(-\frac{2}{x^2} + 3\)[/tex]:
- Because [tex]\(\frac{2}{x^2}\)[/tex] is always positive, [tex]\(-\frac{2}{x^2}\)[/tex] is always negative.
- Therefore, [tex]\(-\frac{2}{x^2} + 3\)[/tex] will always be less than 3 (since we are subtracting a positive quantity from 3).
- As [tex]\(x\)[/tex] approaches [tex]\(\pm\infty\)[/tex], [tex]\(\frac{1}{x^2}\)[/tex] approaches 0 and [tex]\(-\frac{2}{x^2}\)[/tex] approaches 0. Therefore, [tex]\(-\frac{2}{x^2} + 3\)[/tex] will approach 3 but never actually reach 3.
- As [tex]\(x\)[/tex] approaches 0 (from either the positive or negative direction), [tex]\(\frac{1}{x^2}\)[/tex] approaches [tex]\(\infty\)[/tex] and [tex]\(-\frac{2}{x^2}\)[/tex] approaches [tex]\(-\infty\)[/tex], making [tex]\(-\frac{2}{x^2} + 3\)[/tex] become very negative.
Hence, the value of [tex]\(-\frac{2}{x^2} + 3\)[/tex] can be any real number less than 3.
Therefore, the range of the composite function [tex]\((u \circ v)(x)\)[/tex] is:
[tex]\[ (-\infty, 3) \][/tex]
Thus, the correct answer is:
[tex]\[ (-\infty, 3) \][/tex]
