Question

f(x) = 16x² \[-\] 16x + 28 on R ?

Answer 1

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.