Question

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

Answer 1

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(x-1) = x^2-x`

`d/dx(x^2-x)=2x-1`

`2x-1=0=>x=1/2`

This lies on the interval `[0,1]`

So, a local extrema is possible, not guaranteed.

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

`(2x-1)^1`

There will be a local extrema at `x=1/2`

Let's plug in `0`.

`2(0)-1=-1=>`negative slope from `[0,1/2]`

Again, since the multiplicity is odd, there will be a positive slope from `[1/2, 2]`

`:.`There is a local minimum at `x=1/2`