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TLDR: Find where $ f’(x) = 0 $ or is undefined

Critical points are pivotal in understanding the behavior of a function. They help us find local maxima, local minima, and saddle points, which are key to sketching accurate graphs and solving optimization problems.

What are Critical Points?

A critical point of a function $ f(x) $ occurs at any $ x $ -value where either:

1. $ f’(x) = 0 $ (the derivative is zero)
2. $ f’(x) $ is undefined (the derivative doesn’t exist)

In simpler terms, critical points are locations where the function’s slope is either zero or undefined.

Finding Critical Points

The process involves two steps:

1. Find the derivative: Calculate $ f’(x) $ .
2. Solve for critical points: Set $ f’(x) = 0 $ and solve for $ x $ . Also, identify any $ x $ -values where $ f’(x) $ is undefined (e.g., points of discontinuity or sharp corners).

Example:

Let’s find the critical points of the function $ f(x) = x^3 - 3x $ .

1. The derivative is $ f’(x) = 3x^2 - 3 $ .
2. Setting $ f’(x) = 0 $ , we get $ 3x^2 - 3 = 0 $ , which simplifies to $ x^2 = 1 $ . Thus, $ x = 1 $ and $ x = -1 $ are critical points.
3. $ f’(x) $ is defined for all real numbers.

Therefore, the critical points are $ x = 1 $ and $ x = -1 $ .

Classifying Critical Points: The First Derivative Test

The first derivative test helps determine whether a critical point is a local maximum, local minimum, or neither.

1. Test points: Choose test points in the intervals created by the critical points.
2. Evaluate the derivative: Evaluate $ f’(x) $ at each test point.
3. Interpret the sign:
   • If $ f’(x) $ changes from positive to negative, the critical point is a local maximum.
   • If $ f’(x) $ changes from negative to positive, the critical point is a local minimum.
   • If $ f’(x) $ does not change sign, the critical point is neither a local maximum nor a local minimum (it’s a saddle point or an inflection point).

Example (continued):

For $ f(x) = x^3 - 3x $ , the critical points are $ x = 1 $ and $ x = -1 $ .

• For $ x < -1 $ , $ f’(x) > 0 $ .
• For $ -1 < x < 1 $ , $ f’(x) < 0 $ .
• For $ x > 1 $ , $ f’(x) > 0 $ .

Thus, $ x = -1 $ is a local maximum, and $ x = 1 $ is a local minimum.

y = x^3 - 3x

First Derivative Test

Second Derivative Test

Classifying Critical Points: The Second Derivative Test

The second derivative test provides an alternative method for classifying critical points. It’s often easier to use than the first derivative test, but it only works when the second derivative exists.

1. Find the second derivative: Calculate $ f’’(x) $ .
2. Evaluate at critical points: Evaluate $ f’’(x) $ at each critical point.
3. Interpret the sign:
   • If $ f’’(x) < 0 $ , the critical point is a local maximum.
   • If $ f’’(x) > 0 $ , the critical point is a local minimum.
   • If $ f’’(x) = 0 $ , the test is inconclusive (you need to use the first derivative test).

Example (continued):

For $ f(x) = x^3 - 3x $ , $ f’’(x) = 6x $ .

• At $ x = -1 $ , $ f’’(-1) = -6 < 0 $ , so $ x = -1 $ is a local maximum.
• At $ x = 1 $ , $ f’’(1) = 6 > 0 $ , so $ x = 1 $ is a local minimum.

Important Note: Local vs. Global Extrema

Critical points help find local extrema (maxima or minima within a specific interval). To find global extrema (the absolute highest or lowest values over