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This rundown covers using the first derivative to analyze the behavior of a function. We’ll focus on finding Critical Point, increasing/decreasing intervals, and local extrema.
1. Finding Critical Points
Critical points are points in the domain of a function where the derivative is either zero or undefined. These points are potential locations for local maxima, local minima, or neither.
To find critical points:
- Find the first derivative: $ f’(x) $
- Set the derivative equal to zero: $ f’(x) = 0 $ and solve for $ x $ . These are the critical points where the derivative is zero.
- Find where the derivative is undefined: Determine if there are any points in the domain of $ f(x) $ where $ f’(x) $ is undefined (e.g., division by zero, square root of a negative number). These are also critical points.
Example:
Let $ f(x) = x^3 - 3x $ . Then $ f’(x) = 3x^2 - 3 $ .
Setting $ f’(x) = 0 $ , we get $ 3x^2 - 3 = 0 $ , which gives $ x = \pm 1 $ .
$ f’(x) $ is defined for all real numbers, so there are no additional critical points where the derivative is undefined.
Therefore, the critical points are $ x = 1 $ and $ x = -1 $ .
2. Increasing and Decreasing Intervals
The first derivative tells us about the function’s increasing and decreasing behavior:
- ** $ f’(x) > 0 $ :** $ f(x) $ is increasing on this interval.
- ** $ f’(x) < 0 $ :** $ f(x) $ is decreasing on this interval.
To find increasing/decreasing intervals:
- Find the critical points.
- Test intervals: Choose test points in the intervals created by the critical points. Plug these test points into $ f’(x) $ . If $ f’(x) > 0 $ , the function is increasing in that interval; if $ f’(x) < 0 $ , it’s decreasing.
Example (continuing from above):
For $ f(x) = x^3 - 3x $ , the critical points are $ x = -1 $ and $ x = 1 $ .
- Interval $ (-\infty, -1) $ : Test point $ x = -2 $ . $ f’(-2) = 3(-2)^2 - 3 = 9 > 0 $ , so $ f(x) $ is increasing on $ (-\infty, -1) $ .
- Interval $ (-1, 1) $ : Test point $ x = 0 $ . $ f’(0) = -3 < 0 $ , so $ f(x) $ is decreasing on $ (-1, 1) $ .
- Interval $ (1, \infty) $ : Test point $ x = 2 $ . $ f’(2) = 9 > 0 $ , so $ f(x) $ is increasing on $ (1, \infty) $ .
3. Local Extrema (Local Maxima and Minima)
First Derivative test
Local extrema occur at critical points. The first derivative test helps determine whether a Critical Point is a local maximum, local minimum, or neither:
- If $ f’(x) $ changes from positive to negative at a Critical Point $ c $ , then $ f(c) $ is a local maximum.
- If $ f’(x) $ changes from negative to positive at a Critical Point $ c $ , then $ f(c) $ is a local minimum.
- If $ f’(x) $ does not change sign at a Critical Point $ c $ , then $ f(c) $ is neither a local maximum nor a local minimum (it could be a saddle point).
Example (continuing from above):
- At $ x = -1 $ , $ f’(x) $ changes from positive to negative, so $ f(-1) = (-1)^3 - 3(-1) = 2 $ is a local maximum.
- At $ x = 1 $ , $ f’(x) $ changes from negative to positive, so $ f(1) = (1)^3 - 3(1) = -2 $ is a local minimum.
y = x^3 - 3x
Second Derivative Test The second derivative test provides an alternative method to classify critical points. It’s often easier than the first derivative test but has limitations.