# How do you find all local maximum and minimum points given y=x^2-x?

Feb 3, 2018

See explanation.

#### Explanation:

To find the local extreme (maximum or mininmum) of a function $f \left(x\right)$ we have to find its derivative and check where it is zero.

Here we have:

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

$f ' \left(x\right) = 2 x - 1$

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

The point ${x}_{0} = \frac{1}{2}$ can be a critical point of $f \left(x\right)$. To check if it is a critical point we have to find out if the derivative changes sign at ${x}_{0} = \frac{1}{2}$

graph{2x-1 [-2.738, 2.737, -1.37, 1.367]}

As we can see the derivative changes sign from negative to positive, so the point ${x}_{0} = \frac{1}{2}$ is a minimum.

The function $f \left(x\right) = {x}^{2} - x$ has a minimum at ${x}_{0} = \frac{1}{2}$.