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Finding the max/min of a curve

A useful application of calculus is finding maximum or minimum value(s) of a function.

In this graph of the function y = x^3 - 12x+ 2 there is a local maximum (at x= -2 and local minimum (at x= 2).

The blue horizontal line shows that the gradient at these points are zero i.e. f'(x) = 0

Using differentiation:

f(x) = x^3 - 12x+ 2 f'(x) = 3x^2 - 12

To find the max/min points make f'(x) = 0

3x^2 -12 = 0 3x^2=12 x^2=4

There are 2 possible solutions, x = 2 or x = -2

How can we tell which solution is the max or min?

Take the second derivative (i.e. differentiate f'(x) to get f''(x) ).

To test point a:

If f''(a) > 0 , a is a minimum

If f''(a) < 0 , a is a maximum

note that if f''(a) = 0, a is a point of inflection

In our example:

f''(x) = 6x

When x= -2, f''(-2) = -12 to show there is a maximum at x=-2

When x= 2, f''(2) = 12 to show there is a minimum at x=2

To see this working explained more fully see this Khan Academy video.

Further information

  • Khan Academy explains how a study of the gradient on various points of a curve will show places where a relative maximum or minimum exist.
  • Desmos is a free online calculator which can be used to draw graphs to show significant features like  local maximum and minimum.