Differentiation

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  • Topic: First derivative test, Second derivative test, Maxima and minima
  • Pages : 6 (1316 words )
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  • Published : January 25, 2013
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C2 differentiation

Maximum points, minimum points and points of inflection

All 3 types of point are easy enough to spot on a graph:

• Maximum points are the tops of ‘peaks’

• Minimum points are bottoms of ‘troughs’

• Points of inflection are where a curve stops turning ‘left’ and starts turning ‘right’ (or vice versa). An example is the point (0,1) on the curve [pic]+1

Notes
(i) Any point on a curve where the gradient is zero can be called a ‘stationary point’ (which means that stationary points include maximum and minimum turning points and also any points of inflection at which the curve is horizontal, as it is in the example given above)

(ii) If a maximum or minimum are not actually the highest or lowest values that a curve ever reaches, but are just the highest or lowest value on the curve near to that point (as is the case for the maximum and minimum points on a cubic curve) then they may be called a ‘local maximum’ or ‘local minimum’. If on the other hand a curve just has a single maximum or just a single minimum (eg a quadratic function) then the value of y at that point will be the greatest or least value of that function.

The task in C2 is to use differentiation to find where these points are on different curves. In the case of maximum and minimum points, we also have to be able to use differentiation to determine which of the two it is.

Finding Maximum and Minimum points

It is fairly obvious that at these points on a curve the gradient is zero.

At a maximum or a minimum point [pic]

Therefore to find a maximum or minimum point (or points), the procedure will be:

i) get the gradient function by differentiating the equation of the curve, then ii) find the value (or values) of x which make the gradient function equal zero

Example: Find the values of ‘x’ at the 2 turning points on the curve [pic] and then use these to calculate the 2 ‘y’ coordinates. Solution: The gradient function is [pic]
Turning points are where [pic] so we need to solve the equation [pic]
Finding the y coordinates is easy – just substitute the x values back into the original curve: [pic]

[pic]

Example: Find the greatest value of the function [pic] Solution: We note that since this is a quadratic function with a negative x2 term, it will be an upside down ‘U’ shape so will have a single turning point which will be a maximum. The question therefore requires us to find the y coordinate of this turning point.

Using differentiation to determine whether a turning point is a maximum or a minimum

The method used for this can be understood by thinking about how the gradient of a curve will be changing as you travel along the curve and pass through a maximum point: the gradient will be positive as you climb up towards the maximum point, it will decrease to zero as you reach the maximum point and then it will become negative as you leave the maximum point behind you. So if we were to plot the values of the gradient as we travel towards and then through a maximum point, we would be plotting a series of decreasing values – positive just before the turning point, zero at the turning point and negative just after the turning point. This means that the gradient itself is decreasing as a maximum point is passed through. For most maximum points (** see note at bottom of page) this means that the gradient of the gradient graph - referred to as [pic] (“d two y by d x squared”) or [pic] - will be negative at a maximum point. It is this observation which is often called the ‘second derivative...
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