How to Find Turning Points Using Differentiation?

Finding turning points of a function using differentiation involves a systematic process of calculus to identify where the function's gradient changes direction, indicating a local maximum or minimum.

What are Turning Points?

A turning point, also known as a critical point or stationary point, is a point on a function's graph where its gradient (slope) is zero. At these points, the curve momentarily flattens out before changing direction. Specifically:

• Local Maximum: The gradient changes from positive to negative.
• Local Minimum: The gradient changes from negative to positive.

For a polynomial function of degree n, there can be up to (n-1) turning points. For example, a quadratic function (degree 2) has at most one turning point, while a cubic function (degree 3) can have up to two.

Steps to Find Turning Points Using Differentiation

The general method involves two main steps: finding the critical points using the first derivative and then determining their nature (maximum or minimum) using either the second derivative test or the first derivative test.

Step 1: Find the First Derivative and Critical Points

The first step is to differentiate the given function, f(x), to find its first derivative, f'(x). The x-coordinates of any turning points are found by setting this derivative equal to zero and solving for x.

1. Differentiate the function: Calculate f'(x).
2. Set the first derivative to zero: Set f'(x) = 0.
3. Solve for x: The values of x obtained are the x-coordinates of the critical points, which are potential turning points.

Step 2: Determine the Nature of the Turning Points (Local Maxima or Minima)

Once you have the x-coordinates of the critical points, you need to determine whether each point is a local maximum or a local minimum. This can be done using either the second derivative test or the first derivative test.

Method A: Second Derivative Test (Preferred for Simplicity)

1. Calculate the second derivative: Find f''(x).
2. Substitute x values: Substitute each x-value (from Step 1) into f''(x).
   • If f''(x) > 0, the point is a local minimum.
   • If f''(x) < 0, the point is a local maximum.
   • If f''(x) = 0, the test is inconclusive. You must use the first derivative test (Method B) or other means to determine the nature of the point. This often indicates a point of inflection.

Here’s a quick reference for the Second Derivative Test:

Value of f''(x)	Nature of Turning Point
f''(x) > 0	Local Minimum
f''(x) < 0	Local Maximum
f''(x) = 0	Inconclusive

Method B: First Derivative Test (Alternative or when Second Derivative Test is Inconclusive)

1. Choose test points: For each x-value from Step 1, choose one test point slightly to its left and one slightly to its right.
2. Substitute test points into f'(x):
   • If f'(x) changes from positive to negative as x increases through the critical point, it is a local maximum.
   • If f'(x) changes from negative to positive as x increases through the critical point, it is a local minimum.
   • If f'(x) does not change sign (e.g., positive to positive or negative to negative), it is generally a point of inflection, not a turning point.

Step 3: Find the Corresponding y-coordinates

Finally, substitute the x-coordinates of the confirmed turning points back into the original function f(x) to find their corresponding y-coordinates. This gives you the complete (x, y) coordinates of each turning point.

Example: Finding Turning Points for a Cubic Function

Let's find the turning points for the function f(x) = x^3 - 3x^2 + 2.

Step 1: Find the First Derivative and Critical Points

1. Differentiate f(x):
   f'(x) = 3x^2 - 6x

2. Set f'(x) = 0:
   3x^2 - 6x = 0

3. Solve for x:
   3x(x - 2) = 0
   This gives x = 0 or x = 2. These are our critical points.

Step 2: Determine the Nature of the Turning Points (Using Second Derivative Test)

1. Calculate the second derivative:
   f''(x) = 6x - 6

2. Substitute x = 0:
   f''(0) = 6(0) - 6 = -6
   Since f''(0) < 0, x = 0 corresponds to a local maximum.

3. Substitute x = 2:
   f''(2) = 6(2) - 6 = 12 - 6 = 6
   Since f''(2) > 0, x = 2 corresponds to a local minimum.

Step 3: Find the Corresponding y-coordinates

1. For x = 0 (local maximum):
   Substitute x = 0 into the original function f(x) = x^3 - 3x^2 + 2:
   f(0) = (0)^3 - 3(0)^2 + 2 = 2
   The local maximum is at (0, 2).

2. For x = 2 (local minimum):
   Substitute x = 2 into the original function f(x) = x^3 - 3x^2 + 2:
   f(2) = (2)^3 - 3(2)^2 + 2 = 8 - 3(4) + 2 = 8 - 12 + 2 = -2
   The local minimum is at (2, -2).

Therefore, the turning points of the function f(x) = x^3 - 3x^2 + 2 are a local maximum at (0, 2) and a local minimum at (2, -2).