Differentiation – Stationary Points and Nature

A stationary point occurs when:

It can be:

• Maximum
• Minimum
• Saddle point (point of inflexion where gradient = 0)

Method A: Using Gradients (First Derivative Test)

• Check the sign of \( \frac{dy}{dx} \) before and after the stationary point.
• Gradient changes + to − → **maximum**
• Gradient changes − to + → **minimum**
• Gradient does not change sign → **saddle point**

Method B: Using the Second Derivative

Evaluate \( \frac{d^2y}{dx^2} \) at the stationary point.

• Positive \( \Rightarrow \) minimum
• Negative \( \Rightarrow \) maximum
• Zero \( \Rightarrow \) test is inconclusive (must use gradients)

Find and classify the stationary point of \(y = x^3 - 3x\).

Step 1: Differentiate: \[ \frac{dy}{dx} = 3x^2 - 3 \] Step 2: Solve \( \frac{dy}{dx} = 0 \): \[ 3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \] Step 3: Check signs:

• At \(x=-1\): gradient goes + to − → **maximum**
• At \(x=1\): gradient goes − to + → **minimum**

Find and classify the stationary point of \(y = x^4\).

Step 1: \[ \frac{dy}{dx} = 4x^3 \] Stationary when \(4x^3 = 0 \Rightarrow x=0\). Step 2: \[ \frac{d^2y}{dx^2} = 12x^2 \] At \(x=0\): \[ 12(0)^2 = 0 \] Second derivative is 0 → inconclusive. Step 3: Check gradients:

• For \(x<0\): gradient is negative
• For \(x>0\): gradient is positive

Find and classify the stationary point of \(y = \sqrt{x} - x\).

First rewrite: \[ y = x^{1/2} - x \] Differentiate: \[ \frac{dy}{dx} = \frac{1}{2}x^{-1/2} - 1 \] Solve \( \frac{dy}{dx} = 0 \): \[ \frac{1}{2\sqrt{x}} = 1 \Rightarrow \sqrt{x} = \frac{1}{2} \Rightarrow x = \frac{1}{4} \] Second derivative: \[ \frac{d^2y}{dx^2} = -\frac{1}{4}x^{-3/2} \] At \(x=\frac{1}{4}\): \[ -\frac{1}{4}\left(\frac{1}{4}\right)^{-3/2} < 0 \] → **maximum**.

• Always rewrite roots and reciprocals as powers before differentiating.
• If \( \frac{d^2y}{dx^2} = 0 \), you MUST use gradients.
• Show differentiation steps clearly — method marks given.
• Don’t use division to solve \( \frac{dy}{dx}=0 \) if it removes solutions.
• A function can have multiple stationary points — check each one.