https://file.aiquickdraw.com/mathbot/file/1723055051362zolu5btx.jpg

Let's solve the problems presented in the image.

### Problem 9

Given the functions \( f \) and \( g \) on \( (0, \infty) \):

\[ f(x) = 2x^4 - \frac{x^2}{3} \]
\[ g(x) = \frac{1}{x^5} \]

We are asked to apply Lagrange's Mean Value Theorem (LMVT).

LMVT states that if a function \( h(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \in (a, b) \) such that:

\[ h'(c) = \frac{h(b) - h(a)}{b - a} \]

Let's define \( h(x) = f(x) - g(x) \):

\[ h(x) = 2x^4 - \frac{x^2}{3} - \frac{1}{x^5} \]

We need to find where \( h(x) = 0 \), i.e., where \( f(x) = g(x) \).

Setting \( f(x) = g(x) \):

\[ 2x^4 - \frac{x^2}{3} = \frac{1}{x^5} \]

Multiplying through by \( x^5 \):

\[ 2x^9 - \frac{x^7}{3} - 1 = 0 \]

Rewriting:

\[ 6x^9 - x^7 - 3 = 0 \]

This polynomial equation is difficult to solve analytically, but the critical part here is to apply LMVT and understand how many times the function intersects.

Since \( f(x) \) and \( g(x) \) are both continuous and differentiable on \( (0, \infty) \), we analyze their derivatives:

\[ f'(x) = 8x^3 - \frac{2x}{3} \]
\[ g'(x) = -\frac{5}{x^6} \]

For LMVT, we need to find where \( h'(x) = 0 \):

\[ h'(x) = 8x^3 - \frac{2x}{3} + \frac{5}{x^6} \]

Setting \( h'(x) = 0 \):

\[ 8x^3 - \frac{2x}{3} + \frac{5}{x^6} = 0 \]

This equation also does not have a simple analytical solution. However, the behavior of \( h(x) \) indicates multiple crossings due to the polynomial nature and the term \( \frac{1}{x^5} \).

Given these considerations, \( f(x) = g(x) \) for at least one value of \( x \). However, examining the nature of the equations and the degrees of the polynomials, it is likely that \( f(x) = g(x) \) at exactly two points. Hence, the correct answer is:

(C) \( f(x) = g(x) \) for exactly two values of \( x \).

### Problem 10

We are given the limit:

\[ \lim_{x \to 0} \left( \frac{2}{x} - \frac{2}{|x|} \right) \]

This can be simplified by considering the definition of absolute value \( |x| \):

- When \( x > 0 \), \( |x| = x \)
- When \( x < 0 \), \( |x| = -x \)

Consider the limit from the right (\( x \to 0^+ \)):

\[ \lim_{x \to 0^+} \left( \frac{2}{x} - \frac{2}{x} \right) = \lim_{x \to 0^+} 0 = 0 \]

Consider the limit from the left (\( x \to 0^- \)):

\[ \lim_{x \to 0^-} \left( \frac{2}{x} - \frac{2}{-x} \right) = \lim_{x \to 0^-} \left( \frac{2}{x} + \frac{2}{x} \right) = \lim_{x \to 0^-} \frac{4}{x} \]

As \( x \to 0^- \), \( \frac{4}{x} \) approaches \( -\infty \).

Since the left-hand limit and the right-hand limit are not equal, the limit does not exist.

Thus, the correct answer is (D) the limit does not exist.

Do you have any questions or need more details? Here are some related questions to consider:

1. What are the conditions for applying Lagrange's Mean Value Theorem?
2. How do you solve polynomial equations analytically?
3. What is the difference between a polynomial and a rational function?
4. Can you explain the concept of one-sided limits?
5. How does the behavior of a function near zero affect its limit?
6. What are some methods for finding the roots of higher-degree polynomials?
7. How do you determine the continuity of a piecewise function?
8. What role does the absolute value function play in evaluating limits?

**Tip:** When evaluating limits involving absolute values, always consider the behavior of the function from both the left and right sides of the point of interest.

Explore how to apply Lagrange's Mean Value Theorem to solve polynomial equations. This problem involves analyzing functions, derivatives, and polynomial equations using advanced calculus concepts. Suitable for advanced high school or college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation