28: Harder Stationary Points © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules

Harder Stationary Points Module C1 AQA Edexcel OCR MEI/OCR Module C2 "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

Harder Stationary Points The stationary points of a curve are the points where the gradient is zero We may be able to determine the nature of a stationary point just by knowing the shape of a curve. e.g.1 We know the curve has a minimum because the sign of the term ( positive ) tells us that the graph has the following shape

Harder Stationary Points x x Plotting the stationary points and using our knowledge that a cubic is a continuous function ( we can draw it with a single stroke ) means we must get the following: e.g.1 The cubic curve has 2 stationary points. They are and x The y- intercept is also useful

Harder Stationary Points x x x Plotting the stationary points and using our knowledge that a cubic is a continuous function ( we can draw it with a single stroke ) means we must get the following: e.g.1 The cubic curve has 2 stationary points. They are and

Harder Stationary Points Using the 2 nd derivative is usually the easiest method. We may not know the shape of some functions, so we need to determine the nature of the stationary points by another method.

Harder Stationary Points Distinguish between the max and the min. Solution: e.g.2 Calculate the coordinates of the stationary points on the graph of where Multiply by : must be written in the form before we can differentiate For st. pts. this quadratic equation has no linear term so there is no need to factorize

Harder Stationary Points N.B. The maximum has a smaller y -value than the minimum ! To distinguish between the stationary points we need the 2 nd derivative The stationary points are ( 1, 2 ) and (  1,  2) is a min is a max It’s interesting to see what the graph looks like. Calculate y -values at x = 1 and  1 :

Harder Stationary Points is infinite, so x = 0 (the y -axis) is an asymptote So, we now have x x (max) (min) x = 0

Harder Stationary Points “ approaches x “ Also, as “ x approaches infinity “ so “ approaches zero “ x x (max) (min) x = 0 is also an asymptote y = x

Harder Stationary Points x x (max) (min) Asymptote, x = 0 We can now complete the curve.

Harder Stationary Points Exercise 1.Find the stationary points on the curve where Determine the nature of the stationary points. The asymptotes are Ans: is a maximum is a minimum The question didn’t ask for the graph but it looks like this:

Harder Stationary Points

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Harder Stationary Points Distinguish between the max and the min. Solution: e.g.2 Calculate the coordinates of the stationary points on the graph of where Multiply by : must be written in the form before we can differentiate For st. pts. this quadratic equation has no linear term so there is no need to factorize

Harder Stationary Points N.B. The maximum has a smaller y -value than the minimum ! To distinguish between the stationary points we need the 2 nd derivative The stationary points are ( 1, 2 ) and (  1,  2) is a min is a max Calculate y -values at x = 1 and  1 :