Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 16x2 –16x + 28 on R
We have f(x) = 16x2 – 16x + 28 on R
= 16x2 – 16x + 4 + 24
= (4x – 2)2 + 24
Now, (4x – 2)2 ≥ 0 for all x ∈ R
= (4x – 2)2 + 24≥ 24 for all x ∈ R
= f(x) ≥ f
Thus, the minimum value of f(x) is 24 at x =
Hence, f(x) can be made large as possibly by giving difference value to x.
Thus, maximum values does not exist.
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