Concavity And Inflection Points For F(x) = √(3x) Ln(2x) Detailed Analysis
In this article, we will explore how to determine the intervals where the function f(x) = √(3x) ln(2x) is concave up or concave down. Additionally, we will identify any inflection points. Concavity is an essential concept in calculus that describes the curvature of a function's graph. A function is concave up if its graph curves upwards, and concave down if it curves downwards. Inflection points are points where the concavity of the function changes.
Before diving into the specifics of the given function, let's briefly review the key concepts:
- Concave Up: A function f(x) is concave up on an interval if its second derivative, f''(x), is positive on that interval. Geometrically, this means that the graph of f(x) is curving upwards, resembling a smile.
- Concave Down: A function f(x) is concave down on an interval if its second derivative, f''(x), is negative on that interval. Geometrically, the graph of f(x) is curving downwards, resembling a frown.
- Inflection Point: An inflection point is a point on the graph of f(x) where the concavity changes. This occurs when f''(x) changes sign. To find inflection points, we need to find where f''(x) = 0 or is undefined and then check if the concavity changes around these points.
Step-by-Step Guide to Finding Concavity and Inflection Points
To determine the concavity and inflection points for f(x) = √(3x) ln(2x), we will follow these steps:
- Find the First Derivative, f'(x): Use the product rule and chain rule as necessary.
- Find the Second Derivative, f''(x): Differentiate f'(x) to find the second derivative.
- Determine Critical Points for Concavity: Find the values of x for which f''(x) = 0 or f''(x) is undefined. These points are potential inflection points.
- Create a Sign Chart for f''(x): Use the critical points to divide the domain of f(x) into intervals. Pick a test point within each interval and evaluate f''(x) at that point to determine the sign of f''(x) in that interval.
- Determine Intervals of Concavity: Based on the sign chart, identify the intervals where f''(x) > 0 (concave up) and f''(x) < 0 (concave down).
- Identify Inflection Points: Check if the concavity changes at the critical points. If it does, those points are inflection points. Find the corresponding y-values for these points.
Given the function f(x) = √(3x) ln(2x), we need to find its first derivative, f'(x). We will use the product rule, which states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). Here, let u(x) = √(3x) and v(x) = ln(2x).
First, we find the derivatives of u(x) and v(x):
u(x) = √(3x) = (3x)^(1/2)
Using the chain rule, u'(x) = (1/2)(3x)^(-1/2) * 3 = (3/2)(3x)^(-1/2) = 3 / (2√(3x))
v(x) = ln(2x)
Using the chain rule, v'(x) = (1 / (2x)) * 2 = 1/x
Now, we apply the product rule:
f'(x) = u'(x)v(x) + u(x)v'(x)
f'(x) = [3 / (2√(3x))] * ln(2x) + √(3x) * (1/x)
f'(x) = (3 ln(2x)) / (2√(3x)) + √(3x) / x
To simplify, we can rewrite the second term:
√(3x) / x = √(3x) / (√(x) * √(x)) = √3 / √(x)
So, f'(x) = (3 ln(2x)) / (2√(3x)) + √3 / √(x)
To further simplify, we find a common denominator:
f'(x) = (3 ln(2x) + 2√3 * √(3)) / (2√(3x)) = (3 ln(2x) + 6) / (2√(3x))
Therefore, the first derivative is:
f'(x) = (3 ln(2x) + 6) / (2√(3x))
Now, we need to find the second derivative, f''(x), by differentiating f'(x) = (3 ln(2x) + 6) / (2√(3x)). We will use the quotient rule, which states that if f'(x) = u(x) / v(x), then f''(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2. Here, let u(x) = 3 ln(2x) + 6 and v(x) = 2√(3x).
First, we find the derivatives of u(x) and v(x):
u(x) = 3 ln(2x) + 6
u'(x) = 3 * (1 / (2x)) * 2 = 3/x
v(x) = 2√(3x) = 2(3x)^(1/2)
v'(x) = 2 * (1/2) * (3x)^(-1/2) * 3 = 3 / √(3x)
Now, we apply the quotient rule:
f''(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2
f''(x) = [(3/x) * 2√(3x) - (3 ln(2x) + 6) * (3 / √(3x))] / (2√(3x))^2
f''(x) = [6√(3x) / x - (9 ln(2x) + 18) / √(3x)] / (4 * 3x)
f''(x) = [6√(3x) / x - (9 ln(2x) + 18) / √(3x)] / (12x)
To simplify, we find a common denominator for the numerator:
f''(x) = [ (6√(3x) * √(3x) - x(9 ln(2x) + 18)) / (x√(3x)) ] / (12x)
f''(x) = [ 18x - 9x ln(2x) - 18x ] / (12x^2√(3x))
f''(x) = -9x ln(2x) / (12x^2√(3x))
f''(x) = -3 ln(2x) / (4x√(3x))
Therefore, the second derivative is:
f''(x) = -3 ln(2x) / (4x√(3x))
To find the critical points for concavity, we need to find the values of x for which f''(x) = 0 or f''(x) is undefined.
f''(x) = -3 ln(2x) / (4x√(3x))
First, let's find where f''(x) = 0:
-3 ln(2x) = 0
ln(2x) = 0
2x = e^0 = 1
x = 1/2
Now, let's find where f''(x) is undefined. This occurs when the denominator is zero:
4x√(3x) = 0
x = 0
However, we must also consider the domain of the original function, f(x) = √(3x) ln(2x). The natural logarithm ln(2x) is defined only for 2x > 0, which means x > 0. The square root √(3x) is defined for 3x ≥ 0, which means x ≥ 0. Thus, the domain of f(x) is x > 0. Therefore, x = 0 is not in the domain.
So, the only critical point we need to consider is x = 1/2.
We will create a sign chart for f''(x) = -3 ln(2x) / (4x√(3x)) using the critical point x = 1/2 and the domain restriction x > 0. This divides the domain into the interval (0, 1/2) and (1/2, ∞).
| Interval | Test Point | ln(2x) | -3ln(2x) | 4x√(3x) | f''(x) | Concavity |
|---|---|---|---|---|---|---|
| 0 < x < 1/2 | 1/4 | ln(1/2) | > 0 | > 0 | > 0 | Concave Up |
| x > 1/2 | 1 | ln(2) | < 0 | > 0 | < 0 | Concave Down |
Based on the sign chart:
- f''(x) > 0 on the interval (0, 1/2), so f(x) is concave up on this interval.
- f''(x) < 0 on the interval (1/2, ∞), so f(x) is concave down on this interval.
An inflection point occurs where the concavity changes. From the sign chart, we see that the concavity changes at x = 1/2. To find the corresponding y-value, we plug x = 1/2 into the original function:
f(1/2) = √(3 * 1/2) * ln(2 * 1/2) = √(3/2) * ln(1) = √(3/2) * 0 = 0
Thus, the inflection point is (1/2, 0).
For the function f(x) = √(3x) ln(2x):
- The function is concave up on the interval (0, 1/2).
- The function is concave down on the interval (1/2, ∞).
- The inflection point is (1/2, 0).
This detailed analysis provides a comprehensive understanding of the concavity and inflection points of the given function. By following the steps outlined, you can apply these techniques to analyze other functions as well.
