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

Question

Isabella Ackerman · Accepted Answer

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.

Christopher Agresta · Answer

undefined To find the critical points of the function $f(x) = x^2 - 5x + 4$, we first find the derivative $f'(x)$ and then solve for $x$ where $f'(x) = 0$ or is undefined. $f'(x) = 2x - 5$ Setting $f'(x) = 0$ to find critical points: $2x - 5 = 0$ $2x = 5$ $x = \frac{5}{2}$ Now, we need to determine the nature of the critical point at $x = \frac{5}{2}$ by looking at the sign of the derivative around this point. Since $f'(x) = 2x - 5$, we can see that when $x < \frac{5}{2}$, $f'(x) < 0$, and when $x > \frac{5}{2}$, $f'(x) > 0$. This means that at $x = \frac{5}{2}$, the function has a local minimum, making it a stable critical point.