at what point does y=3ln(x) have maximum curvature?
give the exact point, not approximation

To find the point at which the function \( y = 3 \ln(x) \) has maximum curvature, we follow these steps:

### Step 1: Recall the formula for curvature
The curvature \( \kappa(x) \) of a function \( y = f(x) \) is given by the formula:

\[
\kappa(x) = \frac{|f''(x)|}{\left( 1 + (f'(x))^2 \right)^{3/2}}
\]

Where \( f'(x) \) is the first derivative, and \( f''(x) \) is the second derivative of the function.

### Step 2: First derivative of \( y = 3 \ln(x) \)
We start by differentiating \( y = 3 \ln(x) \) to find \( f'(x) \):

\[
f'(x) = \frac{d}{dx}(3 \ln(x)) = \frac{3}{x}
\]

### Step 3: Second derivative of \( y = 3 \ln(x) \)
Now, differentiate \( f'(x) = \frac{3}{x} \) to find \( f''(x) \):

\[
f''(x) = \frac{d}{dx}\left(\frac{3}{x}\right) = -\frac{3}{x^2}
\]

### Step 4: Substituting into the curvature formula
Now substitute \( f'(x) = \frac{3}{x} \) and \( f''(x) = -\frac{3}{x^2} \) into the curvature formula:

\[
\kappa(x) = \frac{|-\frac{3}{x^2}|}{\left( 1 + \left(\frac{3}{x}\right)^2 \right)^{3/2}} = \frac{\frac{3}{x^2}}{\left( 1 + \frac{9}{x^2} \right)^{3/2}}
\]

### Step 5: Maximize the curvature
To find the point of maximum curvature, we need to differentiate \( \kappa(x) \) with respect to \( x \) and set the derivative equal to 0.

Let \( \kappa(x) = \frac{3}{x^2 \left( 1 + \frac{9}{x^2} \right)^{3/2}} \). We will now differentiate this expression using the quotient rule.

Let me calculate this derivative for you.

The derivative of the curvature function \( \kappa(x) \) with respect to \( x \) is:

\[
\kappa'(x) = -\frac{6(1 + \frac{9}{x^2})^{-3/2}}{x^3} + \frac{81(1 + \frac{9}{x^2})^{-5/2}}{x^5}
\]

### Step 6: Solve \( \kappa'(x) = 0 \)
To find the critical points where the curvature is maximized, set this derivative equal to zero and solve for \( x \).

Let me calculate the critical point now.The critical points where the curvature could be maximized are \( x = \pm 2.12132034355964 \). Since \( y = 3 \ln(x) \) is only defined for positive \( x \), the valid critical point is:

\[
x = 2.12132034355964 = \sqrt{9/2}
\]

### Exact point of maximum curvature:
The exact value is \( x = \sqrt{\frac{9}{2}} \).

Would you like more details or clarifications? Here are five related questions to extend your understanding:

1. How does the curvature formula apply to other types of functions, such as polynomials or trigonometric functions?
2. What is the significance of the first and second derivatives in determining curvature?
3. How can you use the curvature formula to analyze the behavior of curves in 3D space?
4. What role does curvature play in fields like physics or engineering?
5. What are the geometric interpretations of critical points in the context of curvature?

**Tip:** When solving for extrema of functions, always ensure the function's domain includes the solution points to avoid invalid results.

At what point does y = 3ln(x) have maximum curvature? Give the exact point, not approximation.

Find the exact point where the function y = 3ln(x) has maximum curvature using calculus. This detailed solution covers differentiation, the curvature formula, and solving critical points for advanced math students.

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