Question

How do you find the critical points for `x^2-5x+4` and state whether it is stable or unstable?

Answer 1

The critical points of a single variable function are the points in which its derivative equals zero. If the second derivative is positive in these points, the points are unstable; while if the second derivative is negative, the points are stable.

This is a polynomial function, so the derivative of each term `ax^n` will be given by `a*n*x^{n-1}`.

So, the derivative of `x^2`, applying the rule with `a=1` and `n=2`, results to be `2x`.
The derivative of `-5x`, applying the rule with `a=-5` and `n=1`, results to be `-5`.
The derivative of a constant is zero.

So, the first derivative of `x^2-5x+4` is `2x-5`, which equals zero if and only if `x=5/2`.

The second derivative is the derivative of the derivative, and we get that the derivative of `2x-5` is `2`, which is of course positive.

So, the (only) critical point `x=5/2` is stable.