Q1 of 137 Page 19

Find the maximum and the minimum values, if any, without using derivatives of the following functions:

f(x) = 4x2 – 4x + 4 on R

f(x) = 4x2 – 4x + 4 on R


= 4x2 – 4x + 1 + 3


= (2x – 1)2 + 3


Since, (2x – 1)2 ≥0


= (2x – 1)2 + 3 ≥3


= f(x) ≥ f


Thus, the minimum value of f(x) is 3 at x =


Since, f(x) can be made large. Therefore maximum value does not exist.


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