If x is real, the minimum value of x2 – 8x + 17 is

Let f(x)= x2 – 8x + 17


Applying the first derivative we get



Applying the sum rule of differentiation, so we get



Applying the derivative,


f' (x)=2x-8


Putting f’(x)=0,we get


2x-8=0


2x=8


x=4


Therefore the minimum value of f(x) at x=4 is given by


f(x)= x2 – 8x + 17


f(4)=42-8(4)+17


f(4)=16-32+17


f(4)=1


Hence if x is real, the minimum value of x2 – 8x + 17 is 1


So the correct option is option C.

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