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Differentiation β€” Stationary Points and Nature 44 problems

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Stationary Points and Their Nature (9709 β€” Year 12)

Stationary points occur where the gradient of a curve is zero. We only use functions that can be written as \(y = kx^n\) (except \(n = -1\)).

1. What is a Stationary Point?

A stationary point occurs when:

\[ \frac{dy}{dx} = 0 \]

It can be:

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

2. How Do We Determine the Nature?

Method A: Using Gradients (First Derivative Test)
  • Check the sign of \( \frac{dy}{dx} \) before and after the stationary point.
  • Gradient changes + to βˆ’ β†’ **maximum**
  • Gradient changes βˆ’ to + β†’ **minimum**
  • Gradient does not change sign β†’ **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)
If \( \frac{d^2y}{dx^2} = 0 \), ALWAYS check gradients. Do not assume.

3. Example 1 β€” Use Gradient (First Derivative Test)

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 βˆ’ β†’ **maximum**
  • At \(x=1\): gradient goes βˆ’ to + β†’ **minimum**
Final:
Maximum at \(x=-1\) and minimum at \(x=1\).

4. Example 2 β€” Use the Second Derivative

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 β†’ inconclusive. Step 3: Check gradients:

  • For \(x<0\): gradient is negative
  • For \(x>0\): gradient is positive
So it is a **minimum** at \(x=0\).

Minimum at \(x=0\). Must use gradient test.

5. Example 3 β€” Roots (Rewrite as Powers)

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 \(x=\frac{1}{4}\): \[ -\frac{1}{4}\left(\frac{1}{4}\right)^{-3/2} < 0 \] β†’ **maximum**.

Maximum at \(x=\frac{1}{4}\).

6. Exam Tips (9709)

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