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Analyzing Functions with the second derivative
Concavity describes the curvature of a function’s graph. A function is concave up if its graph curves upward, like a smile, and concave down if its graph curves downward, like a frown. We determine concavity using the second derivative of the function.
The Second Derivative Test for Concavity
The concavity of a function $ f(x) $ at a point $ x=c $ is determined by the sign of its second derivative, $ f’’(c) $ :
- Concave Up: If $ f’’(c) > 0 $ , then the graph of $ f(x) $ is concave up at $ x=c $ .
- Concave Down: If $ f’’(c) < 0 $ , then the graph of $ f(x) $ is concave down at $ x=c $ .
- Inflection Point: If $ f’’(c) = 0 $ and the concavity changes at $ x=c $ (from concave up to concave down or vice versa), then $ x=c $ is an inflection point. Inflection Points
Important Note: $ f’’(c) = 0 $ does not guarantee an inflection point. The concavity must actually change at $ x=c $ . Consider the function $ f(x) = x^4 $ . $ f’’(x) = 12x^2 $ , and $ f’’(0) = 0 $ , but the function is concave up on both sides of $ x=0 $ , so there’s no inflection point at $ x=0 $ .
Finding Intervals of Concavity
To find the intervals where a function is concave up or concave down, we follow these steps:
- Find the second derivative: Calculate $ f’’(x) $ .
- Find critical points of the second derivative: Solve the equation $ f’’(x) = 0 $ or find where $ f’’(x) $ is undefined. These points are potential boundaries for intervals of concavity.
- Test intervals: Choose test points in the intervals determined by the critical points of $ f’’(x) $ . Evaluate $ f’’(x) $ at each test point.
- Determine concavity: If $ f’’(x) > 0 $ in an interval, the function is concave up in that interval. If $ f’’(x) < 0 $ , the function is concave down.
Example
Let’s consider the function $ f(x) = x^3 - 3x^2 + 2 $ .
- First derivative: $ f’(x) = 3x^2 - 6x $
- Second derivative: $ f’’(x) = 6x - 6 $
- Critical points of $ f’’(x) $ : $ f’’(x) = 0 $ when $ 6x - 6 = 0 $ , which means $ x = 1 $ .
- Test intervals: We test the intervals $ (-\infty, 1) $ and $ (1, \infty) $ .
- For $ x = 0 $ (in $ (-\infty, 1) $ ): $ f’’(0) = -6 < 0 $ , so $ f(x) $ is concave down on $ (-\infty, 1) $ .
- For $ x = 2 $ (in $ (1, \infty) $ ): $ f’’(2) = 6 > 0 $ , so $ f(x) $ is concave up on $ (1, \infty) $ .
- Inflection Point: Since the concavity changes at $ x=1 $ , there is an inflection point at $ x=1 $ .
y = x^3 - 3x^2 + 2
This example demonstrates how to find intervals of concavity and identify inflection points. Remember to always check for changes in concavity when the second derivative is zero. A sign chart can be very helpful in organizing your work.