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Inflection points represent a change in the concavity of a function. Understanding them requires a grasp of both the first and second derivatives.
What is Concavity?
A function is concave up if its graph is shaped like a cup ( $ \cup $ ), and concave down if its graph is shaped like a cap ( $ \cap $ ). More formally:
- Concave Up: The function’s slope is increasing.
- Concave Down: The function’s slope is decreasing.
Analyzing Functions with the first derivative
Analyzing Functions with the second derivative
Finding Inflection Points
Inflection points occur where the concavity of a function changes. This means the second derivative, $ f’’(x) $ , changes sign. To find inflection points, we follow these steps:
- Find the second derivative: $ f’’(x) $
- Find critical points of the second derivative: Set $ f’’(x) = 0 $ or find where $ f’’(x) $ is undefined. These are potential inflection points.
- Analyze the sign of $ f’’(x) $ around the critical points: If the sign of $ f’’(x) $ changes from positive to negative (or vice-versa) as $ x $ passes through a Critical Point, then that point is an inflection point. If the sign does not change, it’s not an inflection point.
- Verify that the point is in the domain of the original function.
Important Note: A Critical Point of the second derivative ( $ f’’(x) = 0 $ or $ f’’(x) $ is undefined) is not automatically an inflection point. The concavity must actually change at that point.
Example
Let’s find the inflection points of the function $ f(x) = x^3 - 3x^2 + 2x $ .
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First derivative: $ f’(x) = 3x^2 - 6x + 2 $
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Second derivative: $ f’’(x) = 6x - 6 $
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Critical points: Set $ f’’(x) = 0 $ : $ 6x - 6 = 0 \implies x = 1 $
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Sign analysis:
- For $ x < 1 $ , $ f’’(x) < 0 $ (concave down)
- For $ x > 1 $ , $ f’’(x) > 0 $ (concave up)
Since the concavity changes from down to up at $ x = 1 $ , there is an inflection point at $ x = 1 $ . To find the $ y $ -coordinate, substitute $ x = 1 $ into the original function: $ f(1) = 1^3 - 3(1)^2 + 2(1) = 0 $ . Therefore, the inflection point is $ (1, 0) $ .
y = x^3 - 3x^2 + 2x
Cases where $ f’’(x) $ is undefined
If the second derivative is undefined at a point, you still need to check for a sign change in the concavity around that point to determine if it’s an inflection point. This often happens with functions involving absolute values or fractional exponents.