https://file.aiquickdraw.com/mathbot/file/1726618292094g7xo6pzl.jpg

We are given the function \( f(x) = \sqrt{3x + 1} \) and asked to evaluate the limit:

\[
\lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}
\]

This limit represents the derivative of \( f(x) \) at \( x = 2 \), as it is in the form of the difference quotient. To solve this, let's proceed step by step.

### Step 1: Find \( f(2) \)

Substitute \( x = 2 \) into \( f(x) \):

\[
f(2) = \sqrt{3(2) + 1} = \sqrt{6 + 1} = \sqrt{7}
\]

### Step 2: Find \( f(2 + h) \)

Now substitute \( 2 + h \) into \( f(x) \):

\[
f(2 + h) = \sqrt{3(2 + h) + 1} = \sqrt{6 + 3h + 1} = \sqrt{7 + 3h}
\]

### Step 3: Set up the difference quotient

The difference quotient becomes:

\[
\frac{f(2 + h) - f(2)}{h} = \frac{\sqrt{7 + 3h} - \sqrt{7}}{h}
\]

### Step 4: Rationalize the numerator

To simplify this expression, multiply the numerator and denominator by the conjugate of the numerator:

\[
\frac{\sqrt{7 + 3h} - \sqrt{7}}{h} \times \frac{\sqrt{7 + 3h} + \sqrt{7}}{\sqrt{7 + 3h} + \sqrt{7}} = \frac{(7 + 3h) - 7}{h(\sqrt{7 + 3h} + \sqrt{7})}
\]

Simplifying the numerator:

\[
\frac{7 + 3h - 7}{h(\sqrt{7 + 3h} + \sqrt{7})} = \frac{3h}{h(\sqrt{7 + 3h} + \sqrt{7})}
\]

### Step 5: Cancel the \( h \)

Cancel out the \( h \) from the numerator and denominator:

\[
\frac{3}{\sqrt{7 + 3h} + \sqrt{7}}
\]

### Step 6: Take the limit as \( h \to 0 \)

As \( h \to 0 \), \( \sqrt{7 + 3h} \to \sqrt{7} \), so the expression becomes:

\[
\frac{3}{\sqrt{7} + \sqrt{7}} = \frac{3}{2\sqrt{7}}
\]

### Final Answer:

\[
\lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} = \frac{3}{2\sqrt{7}}
\]

This is the derivative of \( f(x) = \sqrt{3x + 1} \) at \( x = 2 \).

Would you like more details, or do you have any questions?

Here are 5 related questions you can try:
1. What is the derivative of \( f(x) = \sqrt{3x + 1} \) for any \( x \)?
2. How would you solve a similar limit problem for \( f(x) = x^2 + 1 \)?
3. What is the derivative of \( f(x) = \sqrt{5x + 2} \) at \( x = 1 \)?
4. How do you find the derivative using the first principles for \( f(x) = \frac{1}{x} \)?
5. How does rationalizing the numerator help in limit problems?

**Tip:** When solving limits involving square roots, rationalizing the numerator often simplifies the expression and helps find the limit.

If f(x) = sqrt(3x + 1), evaluate the following limit: lim(h→0) [(f(2 + h) - f(2)) / h]

Learn how to find the derivative of f(x) = sqrt(3x + 1) at x = 2 using limits and the difference quotient. This calculus problem involves key concepts like rationalizing the numerator and applying the limit definition of a derivative. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation