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Differentiation

Differentiation

Turning points: maxima and minima

Turning Points Max Min Extended 8 min read

On this page
  1. In this note
  2. Formula bank
  3. What a turning point is
  4. Finding turning points
  5. Maximum or minimum?
  6. Ways they test it
  7. Mistakes to avoid
  8. Summary
  9. Practice
  10. Can I do this?
  11. Quiz

In this note

Syllabus E2.12: turning points
  • E2.12Find turning points by solving \(\dfrac{dy}{dx}=0\).
  • E2.12Decide whether each one is a maximum or a minimum.

Before you start: the gradient function and solving quadratics by factorising.

Formula bank

Turning pointLearn it\(\dfrac{dy}{dx}=0\)the gradient is 0 at the top of a hill or the bottom of a valley.Use it when "find the turning point(s)".
Max or min (gradient either side)Learn it\(+\ \to\ 0\ \to\ -\) maximum; \(-\ \to\ 0\ \to\ +\) minimumrising then falling = top of a hill.Use it when "determine whether it is a maximum or minimum".
Max or min (quick check)Learn it\(\dfrac{d^{2}y}{dx^{2}}<0\) maximum; \(>0\) minimumdifferentiate \(\dfrac{dy}{dx}\) again and put the x-value in.Use it when the same question, if you prefer it.

What a turning point is

y = x³ − 3x² − 9x + 2-224-20-1010xymax (−1, 7)min (3, −25)
One maximum and one minimum.

At the top of a hill (a maximum) or the bottom of a valley (a minimum) the curve is flat for a moment: the tangent is horizontal, so the gradient is 0.

So turning points are where \(\dfrac{dy}{dx}=0\).

Finding turning points

General methodHow to find the turning points of any curve

  1. Differentiate to get \(\dfrac{dy}{dx}\).
  2. Put \(\dfrac{dy}{dx}=0\) and solve for \(x\) (a quadratic for a cubic curve: usually two answers).
  3. Find each \(y\) from the ORIGINAL equation of the curve.
  4. Write the points as \((x,\ y)\).

Example Find the turning point of \(y=x^{2}-6x+5\).

Set dy/dx = 0, solve, then y from the curve.

  1. \(\dfrac{dy}{dx}=2x-6\)
  2. \(2x-6=0\Rightarrow x=3\)
  3. \(y=9-18+5=-4\)

Answer \((3,\ -4)\)

Try it yourself

1
Turning point of \(y=x^{2}+8x\).
Hint
\(2x+8=0\).
Show answer

Answer \((-4,\ -16)\)

2
Turning point of \(y=10x-x^{2}\).
Show answer

Answer \((5,\ 25)\)

Maximum or minimum?

General methodHow to decide (gradient either side)

  1. Pick an \(x\) a little to the left and a little to the right of the turning point.
  2. Work out \(\dfrac{dy}{dx}\) at both. Only the SIGN matters.
  3. \(+\) then \(-\) (up, then down): maximum. \(-\) then \(+\) (down, then up): minimum.
Turning point\(\dfrac{dy}{dx}\) just left\(\dfrac{dy}{dx}\) just rightSo
\((-1,\ 7)\)at \(x=-2\): \(+15\)at \(x=0\): \(-9\)up then down: maximum
\((3,\ -25)\)at \(x=2\): \(-9\)at \(x=4\): \(+15\)down then up: minimum

For \(y=x^{3}-3x^{2}-9x+2\), \(\dfrac{dy}{dx}=3x^{2}-6x-9\). The table checks both turning points.

Quick check (optional). Differentiate again to get \(\dfrac{d^{2}y}{dx^{2}}\) and put the x-value in: negative means maximum, positive means minimum. Here \(\dfrac{d^{2}y}{dx^{2}}=6x-6\): at \(x=-1\) it is \(-12\) (maximum), at \(x=3\) it is \(12\) (minimum). Either method gets the marks.

Quadratics are easy: \(+x^{2}\) is a U shape, so its one turning point is a minimum; \(-x^{2}\) is an ∩ shape, so it is a maximum.

Example Find the turning point of \(y=3+4x-x^{2}\) and say whether it is a maximum or a minimum.

Set dy/dx = 0, then use the shape.

  1. \(\dfrac{dy}{dx}=4-2x=0\Rightarrow x=2\)
  2. \(y=3+8-4=7\)
  3. \(-x^{2}\): ∩ shape, so a maximum

Answer \((2,\ 7)\), maximum

Ways they test it

Type 1 Find the turning point(s)

"Find the coordinates of the turning points of \(y=\dots\)" (4-5 marks)

Spot it "turning point" or "stationary point".

  1. dy/dx = 0.
  2. Solve for x.
  3. y from the curve.

Type 2 Maximum or minimum?

"Determine whether each turning point is a maximum or a minimum. Show how you decide." (2 marks)

Spot it "determine", "show how you decide".

  1. Gradient either side (or the second derivative).
  2. Say clearly which is which.

Type 3 Greatest or least value (Hard)

"Find the greatest possible area."

Spot it "greatest", "least", "maximum", "minimum" in a word problem.

  1. Write the quantity as a formula in x.
  2. Differentiate and set it equal to 0.
  3. Put that x back to get the greatest/least value.

For example 60 m of fencing makes three sides of a field against a wall; the sides at right angles to the wall are x m. Greatest area?

  1. third side \(60-2x\), so \(A=x(60-2x)=60x-2x^{2}\)
  2. \(\dfrac{dA}{dx}=60-4x=0\Rightarrow x=15\)
  3. \(A=15\times30=450\)

Answer 450 m²

Mistakes to avoid

  • ✗ y = 0 at a turning point✓ \(\dfrac{dy}{dx}=0\)The tangent is flat; the point is usually not on the x-axis.
  • ✗ y found from dy/dx (which gives 0)✓ y from the curve's equationdy/dx is 0 there by definition.
  • ✗ "It is a maximum" with no reason✓ show the gradient signs or the second derivativeThe mark is for the method.

Summary

  • Turning points: solve \(\dfrac{dy}{dx}=0\), then y from the curve.
  • Gradient \(+\to-\): maximum; \(-\to+\): minimum.
  • Or: \(\dfrac{d^{2}y}{dx^{2}}<0\) maximum, \(>0\) minimum.
  • \(+x^{2}\) quadratic: minimum; \(-x^{2}\): maximum.

If you see …

  • "turning point"dy/dx = 0.
  • "maximum or minimum"check the gradient either side.
  • "greatest" or "least" in a storymake a formula, differentiate, set to 0.

Practice

Practice: easy, medium, hard

Easy

1
Find the turning point of \(y=x^{2}-4x+7\).
Hint
\(2x-4=0\).
Show working
  1. \(2x-4=0\Rightarrow x=2\)
  2. \(y=4-8+7=3\)

Answer \((2,\ 3)\)

2
Is the turning point of \(y=x^{2}-4x+7\) a maximum or a minimum?
Hint
Look at the sign of \(x^{2}\).
Show working
  1. \(+x^{2}\): U shape, so a minimum

Answer a minimum (U shape)

3
Find the turning point of \(y=5-6x-x^{2}\).
Show working
  1. \(-6-2x=0\Rightarrow x=-3\)
  2. \(y=5+18-9=14\)

Answer \((-3,\ 14)\)

Medium

1
\(y=x^{3}-12x+1\). Find the x-coordinates of the turning points.
Hint
\(3x^{2}-12=0\).
Guide me

What is it asking? the x-values where dy/dx = 0

What do we know? a cubic

Plan differentiate, set to 0, solve

Show working
  1. \(3x^{2}-12=0\)
  2. \(x^{2}=4\Rightarrow x=\pm2\)

Answer \(x=2\) and \(x=-2\)

2
Find the y-coordinates of those turning points.
Hint
Put x = 2 and x = −2 into the curve.
Guide me

What is it asking? the y-values

What do we know? \(x=\pm2\)

Plan substitute into the curve

Show working
  1. \(y(2)=8-24+1=-15\)
  2. \(y(-2)=-8+24+1=17\)

Answer \(-15\) and \(17\)

3
Which turning point of \(y=x^{3}-12x+1\) is the maximum? Give its coordinates.
Hint
Check the gradient at \(x=-3\) and \(x=0\).
Guide me

What is it asking? the maximum

What do we know? turning points at \(x=\pm2\)

Plan gradient + then − means maximum

Show working
  1. at \(x=-3\): \(27-12=15>0\); at \(x=0\): \(-12<0\)
  2. up then down: maximum at \((-2,\ 17)\)

Answer \((-2,\ 17)\)

Hard

1
Find the turning points of \(y=2x^{3}-3x^{2}-12x+5\) and determine their nature.
Hint
\(6x^{2}-6x-12=6(x-2)(x+1)\).
Guide me

What is it asking? two points and which is which

What do we know? a cubic

Plan dy/dx = 0, y from the curve, then gradient either side or the second derivative

Show working
  1. \(6x^{2}-6x-12=0\Rightarrow x^{2}-x-2=0\Rightarrow x=2,\ -1\)
  2. \(y(2)=16-12-24+5=-15\); \(y(-1)=-2-3+12+5=12\)
  3. \(\tfrac{d^{2}y}{dx^{2}}=12x-6\): at 2 it is 18 (minimum); at −1 it is −18 (maximum)

Answer \((-1,\ 12)\) maximum; \((2,\ -15)\) minimum

PartAnswer / workingMarksGuidance
\(6x^{2}-6x-12=0\)M1
\(x=2,\ -1\)A1
\((2,\ -15)\), \((-1,\ 12)\)A1A1
a correct method to decideM1
\((-1,\ 12)\) max, \((2,\ -15)\) minA1
Total: 6 marks
2
60 m of fencing makes three sides of a rectangle against a wall. The two sides at right angles to the wall are \(x\) m. Show that the area is \(A=60x-2x^{2}\) and find the greatest area.
Hint
The third side is \(60-2x\).
Guide me

What is it asking? the greatest area

What do we know? 3 sides use 60 m

Plan write A in x, then \(\tfrac{dA}{dx}=0\)

Show working
  1. \(A=x(60-2x)=60x-2x^{2}\)
  2. \(\tfrac{dA}{dx}=60-4x=0\Rightarrow x=15\)
  3. \(A=15\times30=450\)

Answer 450 m²

PartAnswer / workingMarksGuidance
\(x(60-2x)\) shownB1
\(60-4x=0\)M1
450A1
Total: 3 marks
3
\(y=x^{3}-3x^{2}-9x+2\). Find the coordinates of the minimum point.
Guide me

What is it asking? the minimum point

What do we know? a cubic with two turning points

Plan find both, then decide which is the minimum

Show working
  1. \(3x^{2}-6x-9=0\Rightarrow x=3,\ -1\)
  2. at \(x=3\): \(\tfrac{d^{2}y}{dx^{2}}=6x-6=12>0\), minimum
  3. \(y=27-27-27+2=-25\)

Answer \((3,\ -25)\)

PartAnswer / workingMarksGuidance
\(3x^{2}-6x-9=0\)M1
\(x=3\)A1
\((3,\ -25)\) with a reasonA1
Total: 3 marks

Can I do this?

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Quiz

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