A stationary point (or turning point) occurs where the derivative equals zero—that is, where the gradient of the curve is zero. At such points, the function changes from increasing to decreasing (maximum) or from decreasing to increasing (minimum).

Finding Stationary Points

Step 1: Differentiate the function to find f'(x).

Step 2: Set f'(x) = 0 and solve for x.

Step 3: Substitute the x-values into the original function to find the corresponding y-coordinates.

Classifying Stationary Points: Second Derivative Test

The second derivative test uses f''(x) to classify each stationary point:

• If f''(x) > 0 at the stationary point → minimum point (the curve is concave up)
• If f''(x) < 0 at the stationary point → maximum point (the curve is concave down)
• If f''(x) = 0 → the test is inconclusive; use the first derivative test

Worked Example

Example: Find and classify the stationary points of y = x³ + 3x² - 9x + 6.

Solution:

Step 1: Find the first derivative: $y' = 3x^2 + 6x - 9$

Step 2: Set y' = 0: $3x^2 + 6x - 9 = 0$ $x^2 + 2x - 3 = 0$ $(x - 1)(x + 3) = 0$ Thus, x = 1 or x = -3

Step 3: Find y-coordinates:

• When x = 1: y = 1³ + 3(1)² - 9(1) + 6 = 1 + 3 - 9 + 6 = 1
• When x = -3: y = (-3)³ + 3(-3)² - 9(-3) + 6 = -27 + 27 + 27 + 6 = 33

Stationary points are (1, 1) and (-3, 33).

Step 4: Classify using the second derivative: $y'' = 6x + 6$

• At x = 1: y''(1) = 6(1) + 6 = 12 > 0 → minimum point (1, 1)
• At x = -3: y''(-3) = 6(-3) + 6 = -12 < 0 → maximum point (-3, 33)

When two or more quantities are related through a function, their rates of change are also related. The chain rule connects these rates:

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

In related rates problems, we differentiate an equation that relates the variables with respect to time t.

Worked Example

Example: A conical funnel with radius 10 cm and height 20 cm is leaking water at 5 cm³/s. How fast is the water level dropping when the depth is 10 cm?

Solution:

Let h = depth, r = radius, V = volume.

From similar triangles: r/h = 10/20 = 1/2, so r = h/2.

Volume of cone: V = (1/3)πr²h = (1/3)π(h/2)²h = (1/12)πh³

Differentiate with respect to time t: $\frac{dV}{dt} = \frac{π}{4}h^2 \frac{dh}{dt}$

Given: dV/dt = -5 cm³/s (water is leaving), h = 10 cm

$-5 = \frac{π}{4}(10)^2 \frac{dh}{dt}$

$-5 = 25π \frac{dh}{dt}$

$\frac{dh}{dt} = -\frac{5}{25π} = -\frac{1}{5π} \text{ cm/s}$

The water level is dropping at 1/(5π) cm/s.

When x changes by a small amount Δx, the corresponding change in y (Δy) can be approximated using the derivative:

$\Delta y \approx \frac{dy}{dx} \Delta x$

This is useful for estimating values without performing complex calculations.

Worked Example

Example: Use differentials to approximate √25.08.

Solution:

Let y = √x. We want to find √(25.08).

Choose x = 25 (where we know √25 = 5) and Δx = 0.08.

The derivative: dy/dx = 1/(2√x) = 1/(2 × 5) = 0.1

Approximate change: $\Delta y \approx (0.1)(0.08) = 0.008$

Therefore: $\sqrt{25.08} \approx 5 + 0.008 = 5.008$

Taylor's Series

Taylor's theorem expresses a function f(x) as an infinite sum of terms calculated from the derivatives at a point a:

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$

Maclaurin's Series

This is the special case when a = 0:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

Common Maclaurin Series Expansions

• e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + ...
• sin x = x - x³/3! + x⁵/5! - x⁷/7! + ...
• cos x = 1 - x²/2! + x⁴/4! - x⁶/6! + ...
• ln(1+x) = x - x²/2 + x³/3 - x⁴/4 + ... (for |x| < 1)

Worked Example

Example: Find the Maclaurin series expansion of ln(1+x) up to the x⁴ term.

Solution:

Let f(x) = ln(1+x)

• f(0) = ln(1) = 0
• f'(x) = 1/(1+x), so f'(0) = 1
• f''(x) = -(1+x)⁻², so f''(0) = -1
• f'''(x) = 2(1+x)⁻³, so f'''(0) = 2
• f⁽⁴⁾(x) = -6(1+x)⁻⁴, so f⁽⁴⁾(0) = -6

Using Maclaurin's formula: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \cdots$

$\ln(1+x) = 0 + 1·x + \frac{(-1)}{2!}x^2 + \frac{2}{3!}x^3 + \frac{(-6)}{4!}x^4 + \cdots$

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$

In Tanzania, these calculus concepts are applied in various practical situations. For example, a small business owner in Dar es Salaam selling mkate wa sinia (rice cake) can use stationary points to determine the selling price that maximizes profit—finding where revenue minus cost is greatest. Similarly, a civil engineer at a construction site in Arusha would use rates of change to calculate how fast water levels rise in a reservoir during rain, or use small change approximations to estimate material costs when dimensions vary slightly from planned values. The series expansions are used in calculator and computer algorithms to compute logarithms and trigonometric functions efficiently.