#### Question

f(x) = 16x^{2} \[-\] 16x + 28 on R ?

#### Solution

Given: *f*(*x*) = 16*x*^{2} − 16*x* + 28

\[\Rightarrow\] f(x) = 4(4x^{2} - 4x + 1) + 24

\[\Rightarrow\] f(x) = 4(2x − 1)^{2} + 24

Now,

4(2*x* − 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.

Is there an error in this question or solution?

Solution for question: F(X) = 16x2 \[-\] 16x + 28 on R ? concept: Graph of Maxima and Minima. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science