Use differentiation to find stationary points of functions

IGCSE Additional Mathematics (0606) – Core Topics Overview

1. Functions (Syllabus 1.1 – 1.8)

f(x)=\sqrt{x},   g(x)=x+3

(f∘g)(x)=√(x+3)   (defined for x≥‑3)
(g∘f)(x)=x+3      (defined for x≥0)
        

Worked Example – Finding an Inverse with a Restricted Domain

Given \(f(x)=x^{2}+2x-3\) with the restriction \(x\ge-1\), find \(f^{-1}(x)\).

1. Write \(y=x^{2}+2x-3\).
2. Complete the square: \(y=(x+1)^{2}-4\;\Rightarrow\;(x+1)^{2}=y+4\).
3. Because \(x\ge-1\), take the positive root: \(x+1=\sqrt{y+4}\).
4. Swap symbols: \(f^{-1}(x)=\sqrt{x+4}-1\), with domain \(x\ge-4\).

2. Quadratic Functions (Syllabus 2.1 – 2.5)

• Standard form: \(f(x)=ax^{2}+bx+c\) (\(aeq0\)).
• Vertex (factor) form: \(f(x)=a(x-h)^{2}+k\) where \((h,k)\) is the vertex.
  • Conversion by completing the square: \(h=-\dfrac{b}{2a},\;k=f(h)\).
• Discriminant: \(\Delta=b^{2}-4ac\).
  • \(\Delta>0\) – two distinct real roots.
  • \(\Delta=0\) – one repeated real root (graph touches the \(x\)-axis).
  • \(\Delta<0\) – no real roots (graph does not intersect the \(x\)-axis).
• Maximum / minimum:
  • If \(a>0\) the parabola opens upwards → vertex is a minimum.
  • If \(a<0\) the parabola opens downwards → vertex is a maximum.
  • Both the completing‑square method and the first‑derivative test give the same result (see Section 7).
• Solving quadratic equations:
  • Factorisation (when possible).
  • Quadratic formula \(\displaystyle x=\frac{-b\pm\sqrt{\Delta}}{2a}\).
  • Completing the square.
• Quadratic inequalities:
  • Find the roots, sketch the parabola, and use the sign of \(a\) to decide where the expression is \(>0\) or \(<0\).

Example – Solving a Quadratic Inequality

Solve \(2x^{2}-5x+2<0\).

1. Factor: \((2x-1)(x-2)<0\).
2. Roots: \(x=\tfrac12\) and \(x=2\).
3. Since the leading coefficient \(2>0\), the parabola opens upwards, so the expression is negative between the roots:
\[ \frac12

Example – Maximum/Minimum via Differentiation

Find the vertex of \(f(x)= -3x^{2}+12x-5\) using calculus.

1. Derivative: \(f'(x)=-6x+12\).
2. Set to zero: \(-6x+12=0\Rightarrow x=2\).
3. Second derivative: \(f''(x)=-6<0\) ⇒ local maximum.
4. Function value: \(f(2)=-3(2)^{2}+12(2)-5=7\).
5. Vertex: \((2,7)\).

3. Factors of Polynomials (Syllabus 3.1 – 3.4)

• Remainder theorem: When \(f(x)\) is divided by \((x-a)\), the remainder is \(f(a)\).
• Factor theorem: \((x-a)\) is a factor of \(f(x)\) iff \(f(a)=0\).
• Factorising cubics (and higher‑degree polynomials):
  1. Search for an integer root using the factor theorem (test factors of the constant term).
  2. Divide by the corresponding linear factor (synthetic or long division).
  3. Factor the remaining quadratic (or repeat the process).

Worked Example – Factorising a Cubic

Factorise \(x^{3}-6x^{2}+11x-6\).

1. Test \(x=1\): \(f(1)=0\) ⇒ \((x-1)\) is a factor.
2. Divide: \(\displaystyle\frac{x^{3}-6x^{2}+11x-6}{x-1}=x^{2}-5x+6\).
3. Factor the quadratic: \((x-2)(x-3)\).
4. Result: \((x-1)(x-2)(x-3)\).

4. Equations, Inequalities & Graphs (Syllabus 4.1 – 4.6)

4.1 Absolute‑value equations

• \(|ax+b|=c\;(c\ge0)\) ⇒ \(ax+b=c\) or \(ax+b=-c\).
• \(|ax+b|=dx+e\) – solve by considering the sign of the right‑hand side, or square both sides after checking for extraneous solutions.

4.2 Absolute‑value inequalities

• \(|ax+b|
• \(|ax+b|>c\) ⇒ \(ax+b-c\).

4.3 Solving linear‑trigonometric equations

For equations of the form \(a\sin x+b\cos x=c\) use the auxiliary‑angle method:

1. Write \(a\sin x+b\cos x=R\sin(x+\alpha)\) where \(R=\sqrt{a^{2}+b^{2}}\) and \(\tan\alpha=\dfrac{b}{a}\).
2. Solve \(\sin(x+\alpha)=\dfrac{c}{R}\) (check that \(|c|\le R\)).
3. Give all solutions in the required interval (usually \(0^\circ\le x<360^\circ\) or \(0\le x<2\pi\)).

4.4 Cubic graphs & inequalities

• Identify turning points (stationary points) using differentiation.
• Sketch the curve, marking \(x\)-intercepts, turning points and end behaviour.
• Use the sketch to decide where \(f(x)>0\) or \(f(x)<0\).

Worked Example – Solve \(|2x-3|=x+1\)

1. Case 1: \(2x-3\ge0\;(x\ge1.5)\) ⇒ \(2x-3=x+1\Rightarrow x=4\) (valid).
2. Case 2: \(2x-3<0\;(x<1.5)\) ⇒ \(-(2x-3)=x+1\Rightarrow -2x+3=x+1\Rightarrow 3x=2\Rightarrow x=\tfrac23\) (valid).
3. Solutions: \(x=\dfrac23,\;4\).

5. Simultaneous Equations (Syllabus 5.1 – 5.4)

• Linear–linear systems: substitution or elimination.
• Linear–quadratic systems:
  1. Express the linear equation for one variable.
  2. Substitute into the quadratic.
  3. Solve the resulting quadratic and back‑substitute.
• Quadratic–quadratic systems:
  1. Subtract the equations to eliminate the squared terms, giving a linear relation.
  2. Use that linear relation in either original equation to obtain a quadratic in one variable.
  3. Solve and back‑substitute.

Worked Example – Solving a Quadratic–Quadratic System

\[ \begin{cases} y= x^{2}+2x-3\\[2mm] y= -x^{2}+4x+1 \end{cases} \]
1. Equate the right‑hand sides: \(x^{2}+2x-3=-x^{2}+4x+1\).
2. Combine: \(2x^{2}-2x-4=0\Rightarrow x^{2}-x-2=0\).
3. Factor: \((x-2)(x+1)=0\) ⇒ \(x=2\) or \(x=-1\).
4. Find \(y\):
   • \(x=2\): \(y=2^{2}+2\cdot2-3=5\).
   • \(x=-1\): \(y=(-1)^{2}+2(-1)-3=-2\).
5. Solutions: \((2,5)\) and \((-1,-2)\).

6. Logarithmic & Exponential Functions (Syllabus 6.1 – 6.5)

• Exponential form: \(y=a^{x}\) (\(a>0,\;aeq1\)).
  • Graph passes through \((0,1)\); horizontal asymptote \(y=0\).
• Logarithmic form: \(y=\log_{a}x\) (inverse of \(a^{x}\)).
  • Domain \(x>0\); vertical asymptote \(x=0\).
• Key laws (valid for \(a,b>0,\;aeq1\)): \[ \begin{aligned} \log_{a}(bc)&=\log_{a}b+\log_{a}c,\\ \log_{a}\!\left(\frac{b}{c}\right)&=\log_{a}b-\log_{a}c,\\ \log_{a}(b^{n})&=n\log_{a}b,\\ a^{\log_{a}b}&=b,\\ \log_{a}b&=\frac{\log_{c}b}{\log_{c}a}\quad(\text{change of base}). \end{aligned} \]
• Solving equations:
  • If the same base appears on both sides, equate the exponents.
  • If different bases appear, take logarithms of both sides (common log or natural log).
  • For equations of the type \(ax=b^{x}\) use trial‑and‑error, graphical methods, or rewrite as \(x\log b=\log a\) when possible.

Example – Solve \(3^{2x-1}=27\)

Write \(27=3^{3}\). Then \(3^{2x-1}=3^{3}\) ⇒ \(2x-1=3\) ⇒ \(x=2\).

7. Calculus – Using Differentiation to Find Stationary Points (Syllabus 7.1 – 7.4)

7.1 What is a stationary point?

A stationary point of \(f(x)\) is a point where the gradient (first derivative) is zero: \(f'(x)=0\). It can be a:

• Local maximum,
• Local minimum, or
• Point of inflection (horizontal tangent).

7.2 General procedure

1. Differentiate the function to obtain \(f'(x)\).
2. Set \(f'(x)=0\) and solve for \(x\) – these are the critical points.
3. Substitute each critical \(x\) into \(f(x)\) to get the corresponding \(y\)-coordinates.
4. Classify each point:
   • Second‑derivative test:
     \(f''(x)>0\) ⇒ local minimum;
     \(f''(x)<0\) ⇒ local maximum.
   • If \(f''(x)=0\), examine the sign change of \(f'(x)\) around the point or use higher‑order derivatives.

7.3 Examples

Example 1 – Quadratic (revisited)

Find the stationary point of \(f(x)=2x^{2}-8x+3\).

1. \(f'(x)=4x-8\).
2. Set to zero: \(4x-8=0\Rightarrow x=2\).
3. Value: \(f(2)=2(2)^{2}-8(2)+3=-5\). Point \((2,-5)\).
4. Second derivative: \(f''(x)=4>0\) ⇒ local minimum.

Example 2 – Cubic

Determine all stationary points of \(g(x)=x^{3}-3x^{2}+2\).

1. \(g'(x)=3x^{2}-6x=3x(x-2)\).
2. Critical points: \(x=0\) and \(x=2\).
3. Values: \(g(0)=2\) → \((0,2)\); \(g(2)=-2\) → \((2,-2)\).
4. Second derivative: \(g''(x)=6x-6\).
   • At \(x=0\): \(g''(0)=-6<0\) ⇒ local maximum.
   • At \(x=2\): \(g''(2)=6>0\) ⇒ local minimum.

Example 3 – Quartic with an inflection point

Find and classify the stationary points of \(h(x)=x^{4}-4x^{3}\).

1. \(h'(x)=4x^{3}-12x^{2}=4x^{2}(x-3)\).
2. Critical points: \(x=0\) (double root) and \(x=3\).
3. Values: \(h(0)=0\) → \((0,0)\); \(h(3)=3^{4}-4\cdot3^{3}=81-108=-27\) → \((3,-27)\).
4. Second derivative: \(h''(x)=12x^{2}-24x=12x(x-2)\).
   • At \(x=0\): \(h''(0)=0\) – inconclusive.
   • Examine \(h'(x)\) around \(x=0\): sign changes from positive (for \(x<0\)) to positive (for \(0
   • At \(x=3\): \(h''(3)=12\cdot3\cdot1=36>0\) ⇒ local minimum.

7.4 Quick checklist for exam questions

Step	What to do
1. Differentiate	Write \(f'(x)\) clearly; simplify if possible.
2. Solve \(f'(x)=0\)	Factorise; use quadratic formula if needed.
3. Find \(y\)	Substitute each critical \(x\) back into the original function.
4. Classify	Compute \(f''(x)\). Apply the sign test; if \(f''=0\) check the sign of \(f'\) on either side or go to the third derivative.
5. State answer	List points as \((x,\;y)\) and label “maximum”, “minimum” or “point of inflection”.

Common pitfalls

• For a quadratic, the vertex found by completing the square and the stationary point found by differentiation are the same – make sure the \(x\)-coordinate matches.
• Never forget to check the domain of the original function before accepting a stationary point (e.g., a rational function may have a vertical asymptote at a critical \(x\)).
• If the second derivative is zero, always examine the sign of the first derivative or go to the next non‑zero derivative; otherwise you may mis‑classify an inflection point as a maximum or minimum.