If $u(x)=-2x^2+3$ And $v(x)=\frac{1}{x}$, What Is The Range Of \$(u \cdot V)(x)$[/tex\]?A. $\left(\frac{1}{3}, 0\right$\] B. $(3, \infty$\] C. $(-\infty, 3$\] D. $(-\infty, +\infty$\]

In this problem, we are given two functions, $u(x)=-2x^2+3$ and $v(x)=\frac{1}{x}$, and we need to find the range of their product, $(u \cdot v)(x)$. To do this, we will first find the product of the two functions and then analyze its behavior to determine its range.

Finding the Product of the Two Functions

To find the product of the two functions, we simply multiply them together:

$$(u \cdot v)(x) = u(x) \cdot v(x) = (-2x^2 + 3) \cdot \frac{1}{x}$$

Using the distributive property, we can rewrite this as:

$$(u \cdot v)(x) = -2x^2 \cdot \frac{1}{x} + 3 \cdot \frac{1}{x}$$

Simplifying, we get:

$$(u \cdot v)(x) = -2x + \frac{3}{x}$$

Analyzing the Behavior of the Product Function

Now that we have found the product function, we need to analyze its behavior to determine its range. To do this, we can start by finding the domain of the function. Since the function is defined as long as $x \neq 0$, the domain is all real numbers except $x = 0$.

Next, we can find the critical points of the function by taking its derivative and setting it equal to zero:

$$\frac{d}{dx}(-2x + \frac{3}{x}) = -2 + \frac{3}{x^2} = 0$$

Solving for $x$, we get:

$$x^2 = \frac{3}{2}$$

$$x = \pm \sqrt{\frac{3}{2}}$$

These are the critical points of the function.

Determining the Range of the Product Function

Now that we have found the critical points, we can use them to determine the range of the function. To do this, we can analyze the behavior of the function on either side of the critical points.

For $x < -\sqrt{\frac{3}{2}}$, the function is decreasing, and for $x > \sqrt{\frac{3}{2}}$, the function is increasing. This means that the function has a minimum value at $x = -\sqrt{\frac{3}{2}}$ and a maximum value at $x = \sqrt{\frac{3}{2}}$.

Since the function is decreasing for $x < -\sqrt{\frac{3}{2}}$ and increasing for $x > \sqrt{\frac{3}{2}}$, the minimum value of the function is the value of the function at $x = -\sqrt{\frac{3}{2}}$, and the maximum value of the function is the value of the function at $x = \sqrt{\frac{3}{2}}$.

Finding the Minimum and Maximum Values of the Function

To find the minimum and maximum values of the function, we can plug in the critical points into the function:

$$(u \cdot v)(-\sqrt{\frac{3}{2}}) = -2(-\sqrt{\frac{3}{2}}) + \frac{3}{-\sqrt{\frac{3}{2}}}$$

Simplifying, we get:

$$(u \cdot v)(-\sqrt{\frac{3}{2}}) = 2\sqrt{\frac{3}{2}} - \frac{3}{\sqrt{\frac{3}{2}}}$$

$$(u \cdot v)(-\sqrt{\frac{3}{2}}) = 2\sqrt{\frac{3}{2}} - 2\sqrt{\frac{3}{2}}$$

$$(u \cdot v)(-\sqrt{\frac{3}{2}}) = 0$$

Similarly, we can find the maximum value of the function by plugging in $x = \sqrt{\frac{3}{2}}$:

$$(u \cdot v)(\sqrt{\frac{3}{2}}) = -2\sqrt{\frac{3}{2}} + \frac{3}{\sqrt{\frac{3}{2}}}$$

Simplifying, we get:

$$(u \cdot v)(\sqrt{\frac{3}{2}}) = -2\sqrt{\frac{3}{2}} + 2\sqrt{\frac{3}{2}}$$

$$(u \cdot v)(\sqrt{\frac{3}{2}}) = 3$$

Conclusion

In conclusion, the range of the product function $(u \cdot v)(x)$ is all real numbers except $0$. This means that the correct answer is:

• D. $(-\infty, +\infty)$

This is because the function has a minimum value of $0$ and a maximum value of $3$, and it is defined for all real numbers except $x = 0$.

Final Answer

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In our previous article, we explored the problem of finding the range of the product of two functions, $u(x)=-2x^2+3$ and $v(x)=\frac{1}{x}$. We found that the range of the product function $(u \cdot v)(x)$ is all real numbers except $0$. In this article, we will answer some common questions related to this problem.

Q: What is the domain of the product function?

A: The domain of the product function is all real numbers except $x = 0$. This is because the function is defined as long as $x \neq 0$.

Q: How do I find the critical points of the product function?

A: To find the critical points of the product function, you need to take its derivative and set it equal to zero. The derivative of the product function is:

$$\frac{d}{dx}(-2x + \frac{3}{x}) = -2 + \frac{3}{x^2}$$

Setting this equal to zero, we get:

$$-2 + \frac{3}{x^2} = 0$$

Solving for $x$, we get:

$$x^2 = \frac{3}{2}$$

$$x = \pm \sqrt{\frac{3}{2}}$$

These are the critical points of the function.

Q: How do I determine the range of the product function?

A: To determine the range of the product function, you need to analyze the behavior of the function on either side of the critical points. For $x < -\sqrt{\frac{3}{2}}$, the function is decreasing, and for $x > \sqrt{\frac{3}{2}}$, the function is increasing. This means that the function has a minimum value at $x = -\sqrt{\frac{3}{2}}$ and a maximum value at $x = \sqrt{\frac{3}{2}}$.

Q: What is the minimum and maximum value of the product function?

A: The minimum value of the product function is the value of the function at $x = -\sqrt{\frac{3}{2}}$, which is $0$. The maximum value of the function is the value of the function at $x = \sqrt{\frac{3}{2}}$, which is $3$.

Q: Why is the range of the product function all real numbers except $0$?

A: The range of the product function is all real numbers except $0$ because the function has a minimum value of $0$ and a maximum value of $3$. This means that the function can take on any value between $0$ and $3$, but it cannot take on the value $0$.

Q: What is the final answer to the problem?

A: The final answer to the problem is $\boxed{D. $(-\infty, +\infty)$}$. This is because the range of the product function is all real numbers except $0$.

Conclusion

In conclusion, the range of the product function $(u \cdot v)(x)$ is all real numbers except $0$. This means that the correct answer is $\boxed{D. $(-\infty, +\infty)$}$. We hope that this Q&A article has helped to clarify any questions you may have had about the problem. If you have any further questions, please don't hesitate to ask.