# How do you find the local extremas for x(x-1) on [0,1]?

Dec 9, 2016

Finding local extremas involves the first derivative being set equal to 0; then finding out how the derivative acts as you plug in values greater or less than those zeros.

Therefore, we can rewrite $x \left(x - 1\right) = {x}^{2} - x$

$\frac{d}{\mathrm{dx}} \left({x}^{2} - x\right) = 2 x - 1$

$2 x - 1 = 0 \implies x = \frac{1}{2}$

This lies on the interval $\left[0 , 1\right]$

So, a local extrema is possible, not guaranteed.

However, since the multiplicity of our function is odd, that is:

${\left(2 x - 1\right)}^{1}$

There will be a local extrema at $x = \frac{1}{2}$

Let's plug in $0$.

$2 \left(0\right) - 1 = - 1 \implies$negative slope from $\left[0 , \frac{1}{2}\right]$

Again, since the multiplicity is odd, there will be a positive slope from $\left[\frac{1}{2} , 2\right]$

$\therefore$There is a local minimum at $x = \frac{1}{2}$