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Question
f(x) = 4x2 + 4 on R .
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Solution
Given: f(x) = 4x2 − 4x + 4
\[\Rightarrow\] f(x) = (4x2 − 4x + 1)+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 .
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