You find the turning points of a function by using differentiation to locate where the gradient is zero.
Understanding Turning Points
A turning point (also known as a stationary point or local extremum) is a point on the graph of a function where the gradient changes from positive to negative or negative to positive. These points represent local maximums or local minimums of the function.
For a polynomial function of degree n, there can be up to (n−1) turning points.
Using Differentiation to Find Turning Points
Differentiation allows you to find the slope or gradient of a function at any given point. At a turning point, the tangent line is horizontal, meaning the gradient is zero.
The key method, as stated in the reference, is: differentiating the function and setting the derivative equal to zero which will then give the x coordinates of any turning points.
Here’s the process broken down:
- Differentiate the Function: Find the derivative of the original function, commonly denoted as f'(x) or dy/dx. This derivative function gives you the gradient of the original function at any point x.
- Set the Derivative to Zero: Solve the equation f'(x) = 0. The solutions to this equation are the x-coordinates of the turning points.
- Find the Corresponding y-coordinates: Substitute the x-coordinates found in step 2 back into the original function f(x) to find the corresponding y-coordinates of the turning points.
The points (x, y) you find are the turning points of the function.
Example: Finding Turning Points
Let's find the turning points for a simple function, such as f(x) = x² - 4x + 3.
- Differentiate f(x):
f'(x) = 2x - 4 - Set f'(x) to Zero:
2x - 4 = 0
2x = 4
x = 2
The x-coordinate of the turning point is 2. - Find the Corresponding y-coordinate: Substitute x = 2 back into the original function f(x):
f(2) = (2)² - 4(2) + 3
f(2) = 4 - 8 + 3
f(2) = -1
The y-coordinate is -1.
The turning point is at (2, -1). For this quadratic function (degree n=2), we expected up to (2-1)=1 turning point, which we found.
| Step | Action | Example (f(x) = x² - 4x + 3) | Result |
|---|---|---|---|
| 1. Differentiate f(x) | Find f'(x) | f'(x) = 2x - 4 | Derivative function |
| 2. Set f'(x) = 0 and solve for x | Find x-coordinates where gradient is zero | 2x - 4 = 0 => x = 2 | x-coordinate(s) of turning point(s) |
| 3. Substitute x into f(x) | Find y-coordinates by plugging x-values back into the original function | f(2) = (2)² - 4(2) + 3 = -1 | y-coordinate(s) of turning point(s) |
By using differentiation and setting the derivative equal to zero, you can precisely locate the points where the function's graph changes direction.
