 ## 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.

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