# How do I find all the critical points of f(x)=(x-1)^2?

Oct 29, 2015

See the explanation.

#### Explanation:

The critical numbers for a function $f$ and the numbers in the domain of $f$, at which either $f \left(x\right)$ does not exist or $f \left(x\right) = 0$.

For $f \left(x\right) = {\left(x - 1\right)}^{2}$, note that the domain is $\left(- \infty , \infty\right)$.

Find $f ' \left(x\right)$
We can use the power and chain rule to get:

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

Or we can expand $f \left(x\right) = {x}^{2} - 2 x + 1$ and differentiate this polynomial:

$f ' \left(x\right) = 2 x - 2$
Find the critical numbers for $f$

Either way, we see that
f'(x) is never undefined and f'(x) = 0$a t$x=1#.

$1$ is in the domain of $f$, so it is a critical numbers for $f$.

The "critical points" are (most commonly), the same as critical numbers.
But some use a variant terminology in which critical points are points on the graph, so the have both $x$ and $y$ coordinates.

If you are in such a situation, you need to find $y$ when $x = 1$ to finish.

$y = f \left(1\right) = 0$, so the point on the graph is $\left(1 , 0\right)$