Question

f(x) = 4x² + 4 on R .

Answer 1

Given: f(x) = 4x² − 4x + 4

\[\Rightarrow\] f(x) = (4x² − 4x + 1)+3

\[\Rightarrow\] f(x) = (2x − 1)² + 3

Now,
(2x − 1)² \[\geq\] 0 for all x \[\in\] R \[\in\]

\[\Rightarrow\] f(x) = (2x − 1)² + 3 \[\geq\] 3 for all x \[\in\] R

\[\Rightarrow\] f(x) \[\geq\] 3 for all x \[\in\] R

The minimum value of f is attained when (x − 1) = 0.
(2x − 1) = 0
⇒ x = \[\frac{1}{2}\]

Thus, the minimum value of f (x) at x = \[\frac{1}{2}\] is 3.

Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.

Hence, function f does not have a maximum value .