Find the derivative of `f(x)=(x-2)^3`
Set the derivative to zero, then solve. `3(x-2)^2=0`
To find the value of y, substitute 2 in `f(x) =(x-2)^3`
The critical point is `(2,0)` .
The domain of `3(x-2)^2=0` will determine where the derivative is undefined.
All real numbers.
If f is a function and p is a point in the domain, p is a critical point for f if the derivative of f at p is 0 or undefined. Here the domain is all real numbers and the derivative is defined for all real numbers so the critical point is (2,0).
In order to check where `f(x) =(x-2)^3 ` is increasing or decreasing look at these intervals
(-infinity, 2) U (2, infinity).
The derivative is always positive on an interval if the function is increasing. Likewise the derivative is negative on an interval if the function is decreasing. The critical points are where the derivative is zero and it is where the derivative changes signs. It is where the function changes from increasing to decreasing or vice versa.
After substituting numbers, one can see the function
increases on (-infinity, 2) since f(x)>0
increases on (2, infinity) since f(x)>0