https://r2.erweima.ai/mathbot/image/1727005100037ixaufsku.jpg

The function \( f(x) \) is defined as a piecewise function:

\[
f(x) = \begin{cases} 
\frac{x \ln(3x + 1)}{e^{x^2} - 1}, & x > 0 \\
3 \cos x + x, & x \leq 0
\end{cases}
\]

You are asked to:

1. Examine the continuity of the function at \( x = 0 \).
2. Compute \( f'(1) \), the derivative of the function at \( x = 1 \).

### Part (a): Examine the continuity of \( f(x) \) at \( x = 0 \)

To check the continuity at \( x = 0 \), we need to check the following conditions:
1. The left-hand limit as \( x \to 0^- \).
2. The right-hand limit as \( x \to 0^+ \).
3. The function value \( f(0) \).

#### Left-hand limit (\( x \to 0^- \)):

For \( x \leq 0 \), the function is given by:
\[
f(x) = 3 \cos x + x
\]
Taking the limit as \( x \to 0^- \):
\[
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (3 \cos x + x) = 3 \cos(0) + 0 = 3
\]

#### Right-hand limit (\( x \to 0^+ \)):

For \( x > 0 \), the function is given by:
\[
f(x) = \frac{x \ln(3x + 1)}{e^{x^2} - 1}
\]
Taking the limit as \( x \to 0^+ \):
\[
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x \ln(3x + 1)}{e^{x^2} - 1}
\]
Using a first-order approximation as \( x \to 0 \), we know:
\[
\ln(3x + 1) \approx 3x \quad \text{and} \quad e^{x^2} - 1 \approx x^2
\]
Thus:
\[
\lim_{x \to 0^+} \frac{x \ln(3x + 1)}{e^{x^2} - 1} \approx \lim_{x \to 0^+} \frac{x \cdot 3x}{x^2} = \lim_{x \to 0^+} 3 = 3
\]

#### Function value at \( x = 0 \):

For \( x = 0 \), the function is defined by the second piece (since \( x = 0 \leq 0 \)):
\[
f(0) = 3 \cos(0) + 0 = 3
\]

Since the left-hand limit, right-hand limit, and function value at \( x = 0 \) are all equal to 3, the function is continuous at \( x = 0 \).

### Part (b): Compute \( f'(1) \)

We are tasked with finding \( f'(1) \). Since \( f(x) \) for \( x > 0 \) is \( \frac{x \ln(3x + 1)}{e^{x^2} - 1} \), we will use the quotient rule to differentiate this function.

Let:
\[
u(x) = x \ln(3x + 1), \quad v(x) = e^{x^2} - 1
\]
Using the quotient rule:
\[
f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{v(x)^2}
\]

We now compute \( u'(x) \) and \( v'(x) \):

1. \( u(x) = x \ln(3x + 1) \)
   Using the product rule:
   \[
   u'(x) = \ln(3x + 1) + \frac{3x}{3x + 1}
   \]

2. \( v(x) = e^{x^2} - 1 \)
   Using the chain rule:
   \[
   v'(x) = 2x e^{x^2}
   \]

Thus, the derivative \( f'(x) \) is:
\[
f'(x) = \frac{\left( \ln(3x + 1) + \frac{3x}{3x + 1} \right) (e^{x^2} - 1) - x \ln(3x + 1) (2x e^{x^2})}{(e^{x^2} - 1)^2}
\]

We now evaluate this at \( x = 1 \). First, compute the values of \( u(1), u'(1), v(1), \) and \( v'(1) \):

- \( u(1) = 1 \ln(3 \cdot 1 + 1) = \ln(4) \)
- \( u'(1) = \ln(4) + \frac{3}{4} \)
- \( v(1) = e^1 - 1 = e - 1 \)
- \( v'(1) = 2 \cdot 1 \cdot e = 2e \)

Substituting these values into the expression for \( f'(1) \):

\[
f'(1) = \frac{\left( \ln(4) +

Câu 3: Cho hàm số f(x) gồm hai phần:
1) f(x) = \frac{x \ln(3x+1)}{e^{x^2} - 1} khi x > 0
2) f(x) = 3 \cos x + x khi x \leq 0
Yêu cầu: (a) Khảo sát sự liên tục của hàm f(x) tại x = 0 (b) Tính đạo hàm f'(1)

This problem explores the continuity of a piecewise function at x=0, involving limits from both sides, and the calculation of the derivative at x=1 using the quotient rule. The function includes logarithms and trigonometric terms. Suitable for advanced high school or early university 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