Given The Function F ( X ) = 3 X 3 2 − 25 X − 1 2 − 125 , X \textgreater 0 F(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125, \quad X \ \textgreater \ 0 F ( X ) = 3 X 2 3 ​ − 25 X − 2 1 ​ − 125 , X \textgreater 0 , It Can Be Shown That F ′ ( X ) = 9 2 X 1 2 + 25 2 X − 3 2 F^{\prime}(x) = \frac{9}{2}x^{\frac{1}{2}} + \frac{25}{2}x^{-\frac{3}{2}} F ′ ( X ) = 2 9 ​ X 2 1 ​ + 2 25 ​ X − 2 3 ​ .The Equation F ( X ) = 0 F(x) = 0 F ( X ) = 0

Solving the Equation $f(x) = 0$ for the Given Function

In this article, we will explore the solution to the equation $f(x) = 0$ for the given function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$, where $x > 0$. To begin, we need to understand the properties of the function and its derivative. The derivative of the function, denoted as $f^{\prime}(x)$, is given by $\frac{9}{2}x^{\frac{1}{2}} + \frac{25}{2}x^{-\frac{3}{2}}$. We will use this information to analyze the behavior of the function and find its critical points.

Understanding the Function and its Derivative

The given function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$ is a polynomial function with a degree of 3.5. This means that the function has a non-integer exponent, which can make it challenging to analyze. However, we can simplify the function by rewriting it in a more manageable form.

We can rewrite the function as $f(x) = 3x^{\frac{3}{2}} - \frac{25}{x^{\frac{1}{2}}} - 125$. This form makes it easier to see the behavior of the function as $x$ approaches 0 or infinity.

The derivative of the function, $f^{\prime}(x)$, is given by $\frac{9}{2}x^{\frac{1}{2}} + \frac{25}{2}x^{-\frac{3}{2}}$. This derivative represents the rate of change of the function with respect to $x$.

Finding Critical Points

To find the critical points of the function, we need to set the derivative equal to 0 and solve for $x$. This will give us the values of $x$ where the function has a local maximum or minimum.

Setting $f^{\prime}(x) = 0$, we get:

$$\frac{9}{2}x^{\frac{1}{2}} + \frac{25}{2}x^{-\frac{3}{2}} = 0$$

Multiplying both sides by $2x^{\frac{3}{2}}$, we get:

$$9x^2 + 25 = 0$$

This equation has no real solutions, which means that there are no critical points for the function.

Analyzing the Behavior of the Function

Since there are no critical points, we can analyze the behavior of the function by looking at its end behavior. As $x$ approaches 0, the function approaches negative infinity. As $x$ approaches infinity, the function approaches positive infinity.

Solving the Equation $f(x) = 0$

To solve the equation $f(x) = 0$, we need to find the values of $x$ that make the function equal to 0. Since the function has no critical points, we can use the fact that the function approaches negative infinity as $x$ approaches 0.

We can rewrite the function as $f(x) = 3x^{\frac{3}{2}} - \frac{25}{x^{\frac{1}{2}}} - 125$. As $x$ approaches 0, the term $\frac{25}{x^{\frac{1}{2}}}$ approaches infinity, while the term $3x^{\frac{3}{2}}$ approaches 0.

Therefore, as $x$ approaches 0, the function approaches negative infinity. This means that the function has no real roots.

In conclusion, we have analyzed the behavior of the function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$ and its derivative. We have found that the function has no critical points and approaches negative infinity as $x$ approaches 0. Therefore, the equation $f(x) = 0$ has no real roots.

• [1] Calculus, 3rd edition, Michael Spivak
• [2] Calculus, 2nd edition, James Stewart

A.1 Derivative of the Function

The derivative of the function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$ is given by:

$$f^{\prime}(x) = \frac{9}{2}x^{\frac{1}{2}} + \frac{25}{2}x^{-\frac{3}{2}}$$

A.2 Critical Points

To find the critical points of the function, we need to set the derivative equal to 0 and solve for $x$. This will give us the values of $x$ where the function has a local maximum or minimum.

Setting $f^{\prime}(x) = 0$, we get:

$$\frac{9}{2}x^{\frac{1}{2}} + \frac{25}{2}x^{-\frac{3}{2}} = 0$$

Multiplying both sides by $2x^{\frac{3}{2}}$, we get:

$$9x^2 + 25 = 0$$

This equation has no real solutions, which means that there are no critical points for the function.
Q&A: Solving the Equation $f(x) = 0$ for the Given Function

In our previous article, we explored the solution to the equation $f(x) = 0$ for the given function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$, where $x > 0$. We analyzed the behavior of the function and its derivative, and found that the function has no critical points and approaches negative infinity as $x$ approaches 0. In this article, we will answer some frequently asked questions about the solution to the equation $f(x) = 0$.

Q: What is the significance of the derivative of the function?

A: The derivative of the function represents the rate of change of the function with respect to $x$. It is used to analyze the behavior of the function and find its critical points.

Q: Why is the function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$ important?

A: The function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$ is important because it represents a real-world problem that can be solved using calculus. The function has a non-integer exponent, which makes it challenging to analyze.

Q: What is the relationship between the function and its derivative?

A: The derivative of the function is used to analyze the behavior of the function. The derivative represents the rate of change of the function with respect to $x$. By analyzing the derivative, we can find the critical points of the function.

Q: Why is the equation $f(x) = 0$ important?

A: The equation $f(x) = 0$ is important because it represents a real-world problem that can be solved using calculus. The equation has no real roots, which means that the function has no critical points.

Q: What is the significance of the fact that the function approaches negative infinity as $x$ approaches 0?

A: The fact that the function approaches negative infinity as $x$ approaches 0 is significant because it means that the function has no real roots. This is because the function approaches negative infinity as $x$ approaches 0, which means that the function has no critical points.

Q: Can the equation $f(x) = 0$ be solved using numerical methods?

A: Yes, the equation $f(x) = 0$ can be solved using numerical methods. However, the function has no real roots, which means that the numerical methods will not be able to find a solution.

Q: What is the relationship between the function and the equation $f(x) = 0$?

A: The function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$ is related to the equation $f(x) = 0$ because the equation represents a real-world problem that can be solved using calculus. The function has a non-integer exponent, which makes it challenging to analyze.

In conclusion, we have answered some frequently asked questions about the solution to the equation $f(x) = 0$ for the given function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$, where $x > 0$. We have analyzed the behavior of the function and its derivative, and found that the function has no critical points and approaches negative infinity as $x$ approaches 0.

• [1] Calculus, 3rd edition, Michael Spivak
• [2] Calculus, 2nd edition, James Stewart

A.1 Derivative of the Function

The derivative of the function $f(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125$ is given by:

$$f^{\prime}(x) = \frac{9}{2}x^{\frac{1}{2}} + \frac{25}{2}x^{-\frac{3}{2}}$$

A.2 Critical Points

To find the critical points of the function, we need to set the derivative equal to 0 and solve for $x$. This will give us the values of $x$ where the function has a local maximum or minimum.

Setting $f^{\prime}(x) = 0$, we get:

$$\frac{9}{2}x^{\frac{1}{2}} + \frac{25}{2}x^{-\frac{3}{2}} = 0$$

Multiplying both sides by $2x^{\frac{3}{2}}$, we get:

$$9x^2 + 25 = 0$$

This equation has no real solutions, which means that there are no critical points for the function.