Concavity & Inflection Points: F(x) = (x^2 - 1) / (2x - 1)
In this article, we'll dive deep into the world of concavity and inflection points. We'll take a close look at the function f(x) = (x^2 - 1) / (2x - 1), determining where it's concave up, concave down, and pinpointing those crucial inflection points. Understanding concavity is vital in calculus as it helps us sketch the graph of a function and analyze its behavior. It tells us whether the rate of change of the function is increasing or decreasing.
Understanding Concavity
Before we jump into the specifics of our function, let's solidify our understanding of concavity. Concavity essentially describes the curvature of a function's graph. Imagine you're driving along the curve: if you're turning your steering wheel to the left, the function is concave up; if you're turning to the right, it's concave down. Mathematically, we determine concavity by examining the second derivative of the function.
- Concave Up: A function f(x) is concave up on an interval if its second derivative, f''(x), is positive on that interval. This means the slope of the tangent line is increasing as x increases. Picture a smile or a cup holding water – that's concave up.
- Concave Down: Conversely, a function f(x) is concave down on an interval if its second derivative, f''(x), is negative on that interval. The slope of the tangent line is decreasing as x increases. Think of a frown or an upside-down cup – that's concave down.
- Inflection Points: These are the points where the concavity of the function changes. They occur where the second derivative, f''(x), is either equal to zero or undefined. Inflection points are crucial for understanding the overall shape and behavior of a function's graph.
Step-by-Step Analysis of f(x) = (x^2 - 1) / (2x - 1)
Now, let's apply these concepts to our specific function, f(x) = (x^2 - 1) / (2x - 1). To determine the intervals of concavity and inflection points, we need to follow these key steps:
-
Find the First Derivative, f'(x): The first derivative tells us about the function's increasing and decreasing intervals, but it's also a necessary stepping stone to finding the second derivative. We'll use the quotient rule here, which states that if f(x) = u(x) / v(x), then f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]^2. In our case, u(x) = x^2 - 1 and v(x) = 2x - 1. So, let's calculate:
- u'(x) = 2x
- v'(x) = 2
Applying the quotient rule:
f'(x) = [(2x)(2x - 1) - (x^2 - 1)(2)] / (2x - 1)^2
Simplifying:
f'(x) = (4x^2 - 2x - 2x^2 + 2) / (2x - 1)^2
f'(x) = (2x^2 - 2x + 2) / (2x - 1)^2
-
Find the Second Derivative, f''(x): The second derivative is the key to unlocking the concavity information. We'll differentiate f'(x) using the quotient rule again. This time, u(x) = 2x^2 - 2x + 2 and v(x) = (2x - 1)^2.
- u'(x) = 4x - 2
- To find v'(x), we'll use the chain rule. Let w(x) = 2x - 1. Then v(x) = w(x)^2, and v'(x) = 2w(x)w'(x) = 2(2x - 1)(2) = 8x - 4.
Applying the quotient rule:
f''(x) = [(4x - 2)(2x - 1)^2 - (2x^2 - 2x + 2)(8x - 4)] / (2x - 1)^4
This looks complex, but we can simplify it. Factoring out a 2 from (4x - 2) and a 4 from (8x - 4), we get:
f''(x) = [2(2x - 1)(2x - 1)^2 - 4(2x^2 - 2x + 2)(2x - 1)] / (2x - 1)^4
Now we can factor out a 2(2x - 1) from the numerator:
f''(x) = 2(2x - 1)[(2x - 1)^2 - 2(2x^2 - 2x + 2)] / (2x - 1)^4
Simplifying further, we can cancel a (2x - 1) term:
f''(x) = 2[(2x - 1)^2 - 2(2x^2 - 2x + 2)] / (2x - 1)^3
Expanding and simplifying the numerator:
f''(x) = 2[4x^2 - 4x + 1 - 4x^2 + 4x - 4] / (2x - 1)^3
f''(x) = 2(-3) / (2x - 1)^3
f''(x) = -6 / (2x - 1)^3
-
Find Critical Points of f''(x): Critical points are where f''(x) = 0 or f''(x) is undefined. These are the potential inflection points. Let's analyze our second derivative:
- f''(x) = -6 / (2x - 1)^3
The numerator is a constant, -6, so f''(x) will never be equal to zero. However, f''(x) is undefined when the denominator is zero:
(2x - 1)^3 = 0
2x - 1 = 0
x = 1/2
So, we have a critical point at x = 1/2. This is where the denominator of the original function is also zero, indicating a vertical asymptote. Therefore, x = 1/2 will not be an inflection point because the function is not defined there.
-
Create a Sign Chart for f''(x): A sign chart helps us visualize where f''(x) is positive and negative, thus indicating the intervals of concavity. We'll use the critical point we found (x = 1/2) to divide the number line into intervals.
Interval Test Value (x) f''(x) = -6 / (2x - 1)^3 Sign of f''(x) Concavity x < 1/2 x = 0 -6 / (-1)^3 = 6 + Concave Up x > 1/2 x = 1 -6 / (1)^3 = -6 - Concave Down -
Determine Intervals of Concavity and Inflection Points: Based on the sign chart:
- Concave Up: The function is concave up on the interval (-∞, 1/2) because f''(x) > 0.
- Concave Down: The function is concave down on the interval (1/2, ∞) because f''(x) < 0.
- Inflection Points: Since the concavity changes at x = 1/2, but the function is undefined there, there are no inflection points for this function.
Conclusion
By carefully analyzing the first and second derivatives of f(x) = (x^2 - 1) / (2x - 1), we've determined the intervals where the function is concave up and concave down. We've also learned that while the concavity changes at x = 1/2, it's not an inflection point because the function is undefined at that location.
Understanding concavity and inflection points is a powerful tool in calculus, allowing us to gain a deeper insight into the behavior and shape of functions. It helps us not only sketch graphs accurately but also solve optimization problems and analyze rates of change.
To further expand your understanding of concavity and related concepts, consider exploring resources like Khan Academy's Calculus section. This website offers comprehensive lessons and practice exercises that can solidify your knowledge.
