How do you find the maximum or minimum of #f(x)=x^2-8x+2#?

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Answer 1 · 2018-07-25T20:17:34

minimum #f(4)=-14#

Given function:

#f(x)=x^2-8x+2#

#d/dxf(x)=d/dx(x^2-8x+2)#

#f'(x)=2x-8#

#f''(x)=2>0#

The given function will be minimum where #f'(x)=0# i.e.

#2x-8=0#

#x=8/2#

#x=4#

hence setting #x=4# in given function we get minimum value

#f(4)=4^2-8(4)+2#

#=-14#