Answer :
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]
