Question

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

Answer 1

See the explanation.

Explanation:

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

For `f(x)=(x-1)^2`, note that the domain is `(-oo,oo)`.

Find `f'(x)`
We can use the power and chain rule to get:

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

Or we can expand `f(x) = x^2-2x+1` and differentiate this polynomial:

`f'(x) = 2x-2`
Find the critical numbers for `f`

Either way, we see that
#f'(x) is never undefined and #f'(x) = 0`at`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(1) = 0`, so the point on the graph is `(1,0)`