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

Mar 29, 2015

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 $a {x}^{n}$ will be given by $a \cdot n \cdot {x}^{n - 1}$.

So, the derivative of ${x}^{2}$, applying the rule with $a = 1$ and $n = 2$, results to be $2 x$.
The derivative of $- 5 x$, 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 ${x}^{2} - 5 x + 4$ is $2 x - 5$, which equals zero if and only if $x = \frac{5}{2}$.

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

So, the (only) critical point $x = \frac{5}{2}$ is stable.