Stationary Points, Increasing and Optimising

Roots and fractions as powers; dy/dx = 0 as a quadratic in x² or √x

In the exam for Edexcel International AS Maths · 14 real past-paper questions below

Stationary points: harder algebra

You can already rewrite roots as powers and spot hidden quadratics. In AS questions, $\frac{d y}{d x} = 0$ often needs that algebra, and sometimes a cubic.

Worked example

1
Find the stationary points of $y = 3 x^{2} - 20 x^{\frac{3}{2}} + 36 x$ , for $x > 0$ .
2
$\frac{d y}{d x} = 6 x - 30 x^{\frac{1}{2}} + 36$ — Power rule: $\frac{3}{2} \times 20 = 30$ , and the power drops to $\frac{1}{2}$ .
3
$6 x - 30 x + 36 = 0$ — Stationary means gradient 0.
4
$x - 5 x + 6 = 0$ — Divide every term by 6.
5
$(x)^{2} - 5 x + 6 = 0$ — Spot it: $x$ is $(x)^{2}$ , so this is a quadratic in $x$ .
6
$(x - 2) (x - 3) = 0$ — Factorise as if $x$ were one letter.
7
$x = 2 \Rightarrow x = 4, x = 3 \Rightarrow x = 9$ — Square each answer to get x.
8
$y = 3 (16) - 20 (8) + 36 (4) = 32$ — At $x = 4$ : $4^{\frac{3}{2}} = (4)^{3} = 8$ . The point is (4, 32).
9
$y = 3 (81) - 20 (27) + 36 (9) = 27$ — At $x = 9$ : $9^{\frac{3}{2}} = 3^{3} = 27$ . The point is (9, 27).

Common mistake

Stopping at $x = 2$ and $x = 3$ . Those are values of $x$ , not x. Square them.

Negative powers in $\frac{d y}{d x}$ ? Multiply every term by the highest power of x underneath. If that leaves a cubic, use the factor theorem.

Worked example

1
Find the stationary point of $y = x^{2} - 4 x + \frac{18}{x}$ , for $x > 0$ .
2
$\frac{d y}{d x} = 2 x - 4 - 18 x^{- 2} = 0$ — Differentiate and set equal to 0.
3
$2 x^{3} - 4 x^{2} - 18 = 0$ — Multiply every term by $x^{2}$ .
4
$x^{3} - 2 x^{2} - 9 = 0$ — Divide by 2.
5
$f (3) = 27 - 18 - 9 = 0$ — Try $x = 1, 2, 3, \dots$ until you get 0. So $(x - 3)$ is a factor.
6
$(x - 3) (x^{2} + x + 3) = 0$ — Divide the cubic by $(x - 3)$ to get the quadratic.
7
$b^{2} - 4 a c = 1 - 12 = - 11 < 0$ — The quadratic has no real roots, so $x = 3$ is the only answer.
8
$y = 9 - 12 + 6 = 3$ — Put $x = 3$ into the curve. The stationary point is (3, 3).

Rule

Typical marks: M1 for differentiating, dM1 for setting $\frac{d y}{d x} = 0$ and solving (the d means it needs the first M), then A1 for x and A1 for y. “Show all stages of your working” means an answer read off a calculator gets no marks.

Common mistake

Multiplying only some terms by $x^{2}$ . Every term on both sides gets multiplied, including the plain numbers.

Key idea

Roots in $\frac{d y}{d x}$ : a quadratic in $x$ . Negative powers: multiply through, then look for a hidden quadratic or a factor. Always go back to x, then find y.

Increasing or decreasing?

The sign of the gradient tells you which way the curve is going as you move right.

Rule

Increasing where $\frac{d y}{d x} > 0$ (going uphill). Decreasing where $\frac{d y}{d x} < 0$ (going downhill). Edexcel accepts ≥ or ≤ here too.

Worked example

1
For $y = x^{3} - 9 x^{2} + 24 x + 1$ , find the range of values of x for which y is increasing.
2
$\frac{d y}{d x} = 3 x^{2} - 18 x + 24$ — Differentiate.
3
$3 x^{2} - 18 x + 24 > 0$ — Increasing means gradient positive.
4
$3 (x - 2) (x - 4) > 0$ — Factorise. The critical values are 2 and 4.
5
$x < 2 or x > 4$ — $\frac{d y}{d x}$ is a U-shaped quadratic. It is above 0 outside its roots.
6
$2 < x < 4$ — Between the roots the gradient is negative, so y is decreasing there.

Example

With roots: $y = x^{2} - 32 x$ , $x > 0$ . Then $\frac{d y}{d x} = 2 x - 16 x^{- \frac{1}{2}}$ . Increasing when $2 x > \frac{16}{x}$ , so $x^{\frac{3}{2}} > 8$ , so $x > 4$ . Multiplying by $x$ is safe because it is positive.

Common mistake

Writing $4 < x < 2$ or $2 > x > 4$ . No number fits these. Sketch the U of $\frac{d y}{d x}$ to see which side you need.

Key idea

Increasing or decreasing is a quadratic inequality on $\frac{d y}{d x}$ . Find where it is 0, sketch it, and read off where it is above or below the axis.

Prove there are no turning points

To prove a curve never turns, show that $\frac{d y}{d x} = 0$ has no real solutions. If the gradient is never 0, the curve is never flat.

Worked example

1
Prove that the curve $y = x^{3} - 3 x^{2} + 5 x - 2$ has no turning points.
2
$\frac{d y}{d x} = 3 x^{2} - 6 x + 5$ — Differentiate.
3
$3 x^{2} - 6 x + 5 = 0$ — A turning point would need this to be true.
4
$b^{2} - 4 a c = (- 6)^{2} - 4 (3) (5)$ — Use the discriminant, with $a = 3$ , $b = - 6$ , $c = 5$ .
5
$= 36 - 60 = - 24 < 0$ — Negative, so the equation has no real roots.
6
$So \frac{d y}{d x} \neq = 0 for any x$ — Conclusion: the curve has no turning points.

Rule

The last mark needs all three: the discriminant value, “ $< 0$ ”, and a sentence such as “so there are no turning points”.

Example

Other way: complete the square. $3 x^{2} - 6 x + 5 = 3 (x - 1)^{2} + 2$ , which is always at least 2. So the gradient is never 0, and the curve is always increasing.

Common mistake

Often part (a) asks you to find unknown constants first. Use those values in $\frac{d y}{d x}$ before the discriminant, or the proof falls apart.

Memory trick

Negative discriminant: no roots, never flat, no turns.

Key idea

“Prove no turning points” means: differentiate, set to 0, show the discriminant is negative, and say what it means.

Optimise with a constraint

A common AS question: a shape has a fixed volume, and you want the least surface area. There are two lengths, so you use the volume to remove one of them.

Key idea

Drinks companies solve this problem: hold the same drink with the least metal. Saving a little metal on every can saves a lot over billions of cans.

Worked example

1
An open-topped tank has a base $2 x$ m by $x$ m and height $h$ m. It holds 36 m³. Show that its area of metal is $A = 2 x^{2} + \frac{108}{x}$ .
2
$2 x \times x \times h = 36$ — Volume = length × width × height. This is the constraint.
3
$h = \frac{18}{x ^{2}}$ — Make $h$ the subject: $2 x^{2} h = 36$ .
4
$A = 2 x^{2} + 2 (2 x h) + 2 (x h)$ — Base, plus two long sides, plus two short sides. No top.
5
$A = 2 x^{2} + 6 x h$ — Collect the side terms: $4 x h + 2 x h = 6 x h$ .
6
$A = 2 x^{2} + 6 x \cdot \frac{18}{x ^{2}} = 2 x^{2} + \frac{108}{x}$ — Swap in $h$ . Now A depends on x only, as required.

Worked example

1
Find the value of x that makes A stationary, and show it gives the minimum.
2
$\frac{d A}{d x} = 4 x - 108 x^{- 2}$ — Write $\frac{108}{x}$ as $108 x^{- 1}$ , then differentiate.
3
$4 x - \frac{108}{x ^{2}} = 0$ — Stationary means $\frac{d A}{d x} = 0$ .
4
$4 x^{3} = 108 \Rightarrow x^{3} = 27 \Rightarrow x = 3$ — Multiply by $x^{2}$ , then divide by 4 and take the cube root.
5
$\frac{d ^{2} A}{d x ^{2}} = 4 + 216 x^{- 3}$ — Differentiate again.
6
$4 + \frac{216}{27} = 12 > 0$ — Positive, so $x = 3$ gives a minimum.
7
$A = 2 (9) + \frac{108}{3} = 54 m^{2}$ — Put $x = 3$ into A, not into $\frac{d A}{d x}$ . The least metal is 54 m².

Rule

“Show that” is marked A1*: the given answer earns the mark only if every line before it is shown and correct. “Justify” or “prove it is a minimum” needs the value or sign of the second derivative and the word “minimum”.

Common mistake

Stopping at $x = 3$ when the question asks for the minimum area. Read what it wants: x, h, or A, and give units and the accuracy asked for.

Key idea

Constraint → one variable → differentiate → set to 0 → second derivative → answer the actual question.

Which method? Read the question

Each AS question type has a signal in its wording.

You see	You do
$x$ and $x$ in $\frac{d y}{d x} = 0$	Quadratic in $x$ ; square the answers
Negative powers in $\frac{d y}{d x} = 0$	Multiply through; hidden quadratic or factor theorem
“y is increasing”	Solve $\frac{d y}{d x} > 0$ and state the range
“No turning points”	Discriminant of $\frac{d y}{d x} = 0$ is negative, plus a sentence
Fixed volume, least area	Remove a variable, show that, $\frac{d A}{d x} = 0$ , second derivative

Key idea

Every line starts with a derivative written in powers of x. Get that right and the rest is algebra you already know.

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