Question

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

A. –1
B. 0
C. 1
D. 2

Answer 1

Let f(x)= x² – 8x + 17

Applying the first derivative 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.