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

1 Answer
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 #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.