IB Mathematics AI SL Local maximum and minimum points MAI Study Notes - New Syllabus

IB Mathematics AI SL Local maximum and minimum points MAI Study Notes

LEARNING OBJECTIVE

Values of x where the gradient of a curve is zero.

Key Concepts:

Gradient Zero & Stationary Points
Identifying Local Maxima & Minima

MAI HL and SL Notes – All topics

Gradient Zero & Stationary Points

When the gradient of a curve is zero, the derivative at that point equals zero:

\( f^\prime(x) = 0 \)

This condition is used to identify stationary points, which are locations on a graph where the curve is flat (horizontal tangent).

Definition:

Stationary Point: A point where the derivative \( f^\prime(x) = 0 \)
Gradient = 0 → Horizontal tangent to the curve

Geometric Interpretation:

At a stationary point, the curve momentarily stops increasing or decreasing.
These points may be: Local Maximum, Local Minimum, or a Point of Inflection.

Example

Find the point on the curve \( f(x) = x^2 – 6x + 10 \) where the gradient is zero.

▶️Answer/Explanation

Step 1: Differentiate the function:

\( f'(x) = 2x – 6 \)

Step 2: Set gradient = 0:

\( 2x – 6 = 0 \Rightarrow x = 3 \)

Step 3: Find the corresponding \( y \)-value:

\( f(3) = 3^2 – 6(3) + 10 = 9 – 18 + 10 = 1 \)

Answer: Gradient is zero at the point (3, 1)

Solving \( f^\prime(x) = 0 \)

To find where the curve is flat (stationary points):

Find the derivative: \( f^\prime(x) \)
Solve the equation \( f^\prime(x) = 0 \) to get values of \( x \)
Substitute the values of \( x \) into \( f(x) \) to get the corresponding \( y \)-values

The result is the coordinates of the stationary points.

Example

Find and classify the stationary point(s) of \( f(x) = x^2 – 4x + 7 \).

▶️Answer/Explanation

Step 1: Differentiate the function:

\( f'(x) = 2x – 4 \)

Step 2: Solve \( f'(x) = 0 \):

\( 2x – 4 = 0 \Rightarrow x = 2 \)

Step 3: Find the \( y \)-value:

\( f(2) = 2^2 – 4(2) + 7 = 4 – 8 + 7 = 3 \)

Step 4: Second derivative test:

\( f”(x) = 2 \) → Since \( f”(2) > 0 \), it’s a local minimum

Answer: Stationary point at (2, 3), a local minimum.

Using Technology to Find Derivatives and Stationary Points

Students should be able to:

Use technology (e.g., GDC or software) to generate the derivative \( f'(x) \) given a function \( f(x) \).
Solve \( f'(x) = 0 \) to find stationary points.

Example

Use technology to find the derivative of the function \( f(x) = x^3 – 6x^2 + 9x + 2 \), and solve the equation \( f'(x) = 0 \).

▶️Answer/Explanation

Step 1: On the GDC, enter the function into the graph menu:
f(x) = x^3 - 6x^2 + 9x + 2
Step 2: Use the numerical derivative function:
nDeriv(x^3 - 6x^2 + 9x + 2, x, x)
Step 3: Graph the derivative:
f '(x) = 3x^2 - 12x + 9
Step 4: Use the Calc → zero feature to find where the derivative is zero:
- First zero at \( x = 1 \)
- Second zero at \( x = 3 \)

Final Answer:

\( f'(x) = 3x^2 – 12x + 9 \)
Solutions to \( f'(x) = 0 \): \( x = 1 \), \( x = 3 \)

Identifying Local Maxima & Minima

Definition:

Local Maximum: A peak point where \( f^\prime(x) = 0 \) and the curve turns from increasing to decreasing
Local Minimum: A valley point where \( f^\prime(x) = 0 \) and the curve turns from decreasing to increasing

First Derivative Test:

Check the sign of \( f^\prime(x) \) before and after the stationary point:

If \( f^\prime(x) \) changes from positive to negative → Local Maximum
If \( f^\prime(x) \) changes from negative to positive → Local Minimum

Second Derivative Test:

If \( f^{”}(x) \) exists at the stationary point \( x = a \):

If \( f^{”}(a) > 0 \) → Local Minimum
If \( f^{”}(a) < 0 \) → Local Maximum
If \( f^{”}(a) = 0 \) → Inconclusive, test using first derivative or graph behavior