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F ( X )=4 X 2 -4 X + 4 on R . - Mathematics

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प्रश्न

f(x) = 4x2 + 4 on R .

योग

उत्तर

Given: f(x) = 4x2 − 4x + 4

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

\[\Rightarrow\] f(x) = (2x − 1)2 + 3

Now,
(2x − 1)\[\geq\] 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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अध्याय 18: Maxima and Minima - Exercise 18.1 [पृष्ठ ७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 18 Maxima and Minima
Exercise 18.1 | Q 1 | पृष्ठ ७

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