If the relationship between 'x' and 'y' is non-linear, i.e. not a straight line, 'y' will probably have a maximum and/or a minimum value at some unknown value of 'x'

This maximum and/or minimum value for 'y' will occur when the slope of the curve describing the relationship between 'x' and 'y' is zero.

So instead of performing lots of calculations to try and find out where the maximum value for 'y' occurs, all you have to do is find the first and, if one exists, the second derivative of the formula describing the relationship between 'x' and 'y' and from a few rules (listed below), you can find out everything you need to know.

A function of 'x' is simply a generic term for describing the relationship between 'x' and 'y'. That is a value for 'y' is dependant upon the way 'x' is modified;

i.e. 2.x, x³, Sin(x), etc.

ƒ(x) is simply a short-hand method for saying; a function of 'x'

ƒ'(x) is simply a short-hand method for saying; the first derivative of the function of 'x'

i.e. 2, 3x², Cos(x), etc.

ƒ''(x) is simply a short-hand method for saying; the second derivative of the function of 'x'

i.e. 0, 6x, -Sin(x), etc.

The function of x; ƒ(x) = (x – 1).(x – 2).(x – 3) is the same as; y = (x – 1).(x – 2).(x – 3)

i.e. y = x³ – 6x² + 11x – 6

Its first derivative; ƒ'(x) is: dy/dx = 3x² - 12x + 11

Its second derivative; ƒ''(x) is: d²y/dx² = 6x - 12

Fig 1. y = x³ – 6x² + 11x – 6 and its curves

When dy/dx = 0 the slope of the original curve is also 0 and 'y' must be either a maximum or minimum value⁽¹⁾

3x² - 12x + 11 = 0 when x = 2.577350269 and 1.422649731 (from -b±(b²-4.a.c))^{½} / 2.a)

Putting these values back into y = x³ – 6x² + 11x – 6, you get;

y = 0.384900179 when x = 1.422649731 {we'll call this point 'P'}

y = -0.384900179 when x = 2.577350269 {we'll call this point 'Q'}

When x = 1.422649731, the second derivative; d²y/dx² is -3.464101615 (6x - 12), which is negative so the curve at this point ('P') is a maximum value.

When x = 2.577350269, the second derivative; d²y/dx² is 3.464101615 (6x - 12), which is positive so the curve at this point ('Q') is a minimum value.

And the point of inflection (where the line crosses the 'x' axis) is when d²y/dx² = 0

Which is when 6x - 12 = 0, i.e. when x = 2

Therefore, simply by finding the 1^{st} and 2^{nd} derivatives of 'x' you can find out all the limits of the formula, the shape of the curve and where it sits on the x-y axes.

Maxima & Minima |
If y increases as x increases dy/dx is positive |

Turning point in curve |
dy/dx = 0 |

Maximum |
ƒ(x) increasing before, decreasing after (turning point) |

Minimum |
ƒ(x) decreasing before, increasing after (turning point) |

Turning point |
ƒ(x) changing from concave up to concave down or visa versa |

- This statement excludes special eventualities such as when x=0 or x'=0 or x''=0 etc.

You will find further reading on this subject in reference publications^{(19)}