Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To determine the range of [tex]\((u \circ v)(x)\)[/tex], we need to understand what [tex]\((u \circ v)(x)\)[/tex] means in this context. The notation [tex]\((u \circ v)(x)\)[/tex] signifies the composition of the function [tex]\(u\)[/tex] with the function [tex]\(v\)[/tex]. This means we first apply [tex]\(v(x)\)[/tex] and then apply [tex]\(u\)[/tex] to the result of [tex]\(v(x)\)[/tex]. Mathematically, this is expressed as [tex]\(u(v(x))\)[/tex].
Given the functions:
[tex]\[ u(x) = -2x^2 + 3 \][/tex]
[tex]\[ v(x) = \frac{1}{x} \][/tex]
We need to find [tex]\(u(v(x))\)[/tex].
Firstly, substitute [tex]\(v(x)\)[/tex] into the function [tex]\(u(x)\)[/tex]:
[tex]\[ u(v(x)) = u\left(\frac{1}{x}\right) \][/tex]
Now 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]
Simplify [tex]\(\left(\frac{1}{x}\right)^2\)[/tex]:
[tex]\[ u\left(\frac{1}{x}\right) = -2 \cdot \frac{1}{x^2} + 3 \][/tex]
[tex]\[ u\left(\frac{1}{x}\right) = -\frac{2}{x^2} + 3 \][/tex]
Next, we analyze the expression [tex]\(-\frac{2}{x^2} + 3\)[/tex] to determine its range. The term [tex]\(-\frac{2}{x^2}\)[/tex] is always non-positive because [tex]\(x^2\)[/tex] is always positive for all [tex]\(x \neq 0\)[/tex]. As [tex]\(x\)[/tex] approaches [tex]\(\pm \infty\)[/tex], [tex]\(-\frac{2}{x^2}\)[/tex] approaches 0.
This means that:
[tex]\[ -\frac{2}{x^2} \leq 0 \][/tex]
Thus,
[tex]\[ -\frac{2}{x^2} + 3 \leq 3 \][/tex]
The maximum value of the expression is 3, which occurs as [tex]\(x\)[/tex] approaches [tex]\(\pm \infty\)[/tex]. Since [tex]\(-\frac{2}{x^2}\)[/tex] can be any negative value, [tex]\(-\frac{2}{x^2} + 3\)[/tex] can take on any value less than or equal to 3.
Therefore, the expression [tex]\(-\frac{2}{x^2} + 3\)[/tex] can cover all values from [tex]\(-\infty\)[/tex] to 3.
So, the range of [tex]\((u \circ v)(x)\)[/tex] is:
[tex]\[ (-\infty, 3) \][/tex]
Given the functions:
[tex]\[ u(x) = -2x^2 + 3 \][/tex]
[tex]\[ v(x) = \frac{1}{x} \][/tex]
We need to find [tex]\(u(v(x))\)[/tex].
Firstly, substitute [tex]\(v(x)\)[/tex] into the function [tex]\(u(x)\)[/tex]:
[tex]\[ u(v(x)) = u\left(\frac{1}{x}\right) \][/tex]
Now 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]
Simplify [tex]\(\left(\frac{1}{x}\right)^2\)[/tex]:
[tex]\[ u\left(\frac{1}{x}\right) = -2 \cdot \frac{1}{x^2} + 3 \][/tex]
[tex]\[ u\left(\frac{1}{x}\right) = -\frac{2}{x^2} + 3 \][/tex]
Next, we analyze the expression [tex]\(-\frac{2}{x^2} + 3\)[/tex] to determine its range. The term [tex]\(-\frac{2}{x^2}\)[/tex] is always non-positive because [tex]\(x^2\)[/tex] is always positive for all [tex]\(x \neq 0\)[/tex]. As [tex]\(x\)[/tex] approaches [tex]\(\pm \infty\)[/tex], [tex]\(-\frac{2}{x^2}\)[/tex] approaches 0.
This means that:
[tex]\[ -\frac{2}{x^2} \leq 0 \][/tex]
Thus,
[tex]\[ -\frac{2}{x^2} + 3 \leq 3 \][/tex]
The maximum value of the expression is 3, which occurs as [tex]\(x\)[/tex] approaches [tex]\(\pm \infty\)[/tex]. Since [tex]\(-\frac{2}{x^2}\)[/tex] can be any negative value, [tex]\(-\frac{2}{x^2} + 3\)[/tex] can take on any value less than or equal to 3.
Therefore, the expression [tex]\(-\frac{2}{x^2} + 3\)[/tex] can cover all values from [tex]\(-\infty\)[/tex] to 3.
So, the range of [tex]\((u \circ v)(x)\)[/tex] is:
[tex]\[ (-\infty, 3) \][/tex]
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.