Differentiation (4)- maxima and minimaFinding and Identifying Maxima and minima

Maxima, minima and stationary pointsyxxBDB is a local maximum

D is a local minimum

Points where dy/dx=0 are call stationary points

B & D are also known as turning points(where they gradient changes from +ve to -ve)Terminology

Finding Stationary Points

Examplef(x)= x3 - 12x + 1Find stationary points and determine their typesf(x)= 3x2 - 12Stationary points occur when gradient [derivative] is 03x2 - 12 = 03x2 = 12x2 = 4x = 2 or x = -2When x = -2Pick point to left(x = -2.1)f`(x) = 3x(-2.1)2 -12 = 1.23 [+ve]

Pick point to right(x = -1.9)f`(x) = 3x(-1.9)2 -12 = -1.17 [-ve]

When x = +2Pick point to right(x = 2.1)f`(x) = 3x(2.1)2 -12 = 1.23 [+ve]

Pick point to left(x = 1.9)f`(x) = 3x(1.9)2 -12 = -1.17 [-ve]

Example (continued)f(x)= x3 - 12x + 1 and sketch the graphf(x)= 3x2 - 12Stationary pointsWhen x = -2When x = +2y = (-2)3 -12(-2) +1 = -8 + 24 + 1 = 17

Maximum at (-2,17)y = (2)3 -12(2) +1 = 8 - 24 + 1 = -15

Minimum at (2,-15)When x=0[y-axis]

y = 03 -12x0 +1 = 1

(0,1)

Example (continued)f(x)= x3 - 12x + 1XYTryPage 125Ex AQ1

The second derivativef(x)= x3 - 12x + 1xf(x)-ve at maximum+ve at minimum

The second derivativeKEY POINTSIt is used at a simple way of telling if a stationary point is a maximum or a minimumIf is negative then the stationary point is a maximum If is positive then the stationary point is a minimum

The second derivative - examplef(x) = 1/3x3 - 2x2 + 3x +1Find stationary points and determine their naturef(x) = 2x - 4Stationary points when f(x)=0x2 - 4x + 3 = 0(x - 3)(x - 1) = 0x = 3 and x = 1are maxima or minimaf(x) = x2 - 4x + 3Look at the second derivative to determine their natureWhen x=3f(x) = 2x3 - 4 = 2which is positive hence a minimumWhen x=1f(x) = 2x1 - 4 = -2which is negative hence a maximum

The second derivative - ex C, 4ef(x) = 4x3 - 9x2 - 30x +1Find stationary points and determine their naturef(x) = 24x - 18Stationary points when f(x)=012x2 - 18x - 30 = 02x2 - 3x - 5 = 0(2x - 5)(x + 1) = 0x = 2.5 and x = -1are maxima or minimaf(x) = 12x2 - 18x - 30Look at the second derivative to determine their natureWhen x=2.5f(x) = 24x2.5 - 18 = 42which is positive hence a minimumWhen x=-1f(x) = 24x-1 - 18 = -42which is negative hence a maximum

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