# What is the difference between 'relative maximum(or minimum)' and 'absolute maximum(or minimum)' in functions?

## The terms and definitions were on our textbook and asked my friend about it but she didn't know too. The topic was functions in a graph in Algebra and I came across those four terms.

Mar 17, 2018

A relative maximum or minimum occurs at turning points on the curve where as the absolute minimum and maximum are the appropriate values over the entire domain of the function.

In other words the absolute minimum and maximum are bounded by the domain of the function.

Example:

Consider the Function:

$y = {x}^{4} - 8 {x}^{3} + 22 {x}^{2} - 24 x$

We can find the relative minima and maxima (turning points) by looking for coordinates where the first derivative vanishes:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} - 24 {x}^{2} + 44 x - 24$

The derivative vanishes when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$, ie when

$4 {x}^{3} - 24 {x}^{2} + 44 x - 24 = 0$
$\implies {x}^{3} - 6 {x}^{2} + 11 x - 6 = 0$
$\implies \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) = 0$
$\implies x = 1 , 2 , 3$

And to determine the nature of the turning points we consider the second derivative:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 12 {x}^{2} - 48 x + 44$
$\left(1 , - 9\right) \implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0 \implies \min$
$\left(2 , - 8\right) \implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} < 0 \implies \max$
$\left(3 , - 9\right) \implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} > 0 \implies \min$

And we can plot the graph to verify our findings
graph{y=x^4-8x^3+22x^2-24x [-3, 6, -11, 5]}

So we have:

• Relative minimum of $- 9$ occuring at $x = 1 , 3$
• Relative maximum of $- 8$ occuring at $x = 2$

Over the entire domain as we approach $x = \pm \infty$ the function increases without bound. Subsequently:

• Absolute minimum is also the local minimum, ie $- 9$
• Absolute maximum is unbounded, ie $\infty$