Algebra and calculus

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 is 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.

Learning Hub

Learning Hub team
Workshops
Drop in booth
Learning leaders
Pass Leaders
Business library guides

Life on campus
Careers centre
Student support
He Tuākana
AskAuckland student centre