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प्रश्न
f(x) = 16x2 \[-\] 16x + 28 on R ?
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उत्तर
Given: f(x) = 16x2 − 16x + 28
\[\Rightarrow\] f(x) = 4(4x2 - 4x + 1) + 24
\[\Rightarrow\] f(x) = 4(2x − 1)2 + 24
Now,
4(2x − 1)2 \[\geq\] 0 for all x \[\in\] R
\[\Rightarrow\] f(x) = 4(2x − 1)2 + 24 \[\geq\] 24 for all x \[\in\] R
\[\Rightarrow\] f(x)\[\geq\] 24 for all x \[\in\] R.
The minimum value of f is attained when (2x − 1) = 0.
(2x − 1) = 0
⇒ x = \[\frac{1}{2}\]
Therefore, the minimum value of f at x =\[\frac{1}{2}\] is 24.
Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.
Hence, the function f does not have a maximum value.
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