What are the values and types of the critical points, if any, of #f(x)=x^2-5x+4#?

Question

Cameron Alexander · Accepted Answer

A Minimum point #(5/2, -9/4)# #f(x)=x^2-5x+4# First derivative #f' (x)=2x -5# Equate the #f' (x) = 0# #2x - 5=0# solve for #x# #x=5/2# solve for #y=x^2-5(x)+4# #y=(5/2)^2-5(5/2)+4# #y=25/4-25/2+4# #y=(25-50+16)/4# #y=-9/4#

Christopher Allers · Answer

undefined The critical points of $ f(x) = x^2 - 5x + 4 $ occur where the derivative is zero or undefined.

First, find the derivative:
$$ f'(x) = 2x - 5 $$

Next, set the derivative equal to zero and solve for $ x $ to find the critical points:
$$ 2x - 5 = 0 $$
$$ x = \frac{5}{2} $$

This critical point corresponds to a minimum since the second derivative is positive:
$$ f''(x) = 2 $$

Therefore, the critical point is $ \left(\frac{5}{2}, -\frac{9}{4}\right) $.

There are no other critical points, as the derivative is defined everywhere and does not equal zero for any other value of $ x $.