Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
