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

1 Answer
KillerBunny
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.

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