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

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Answer 1 · 2016-01-12T08:06:32

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#