The minimum value that f(x)=4 x^2-4 x+11+sin 3 π x attains, is

We have,

$f(x)=4 x^2-4 x+11+\sin 3 \pi x$

$\Rightarrow f(x)=4\left(x-\frac{1}{2}\right)^2+10+\sin 3 \pi x$

We observe that $4\left(x-\frac{1}{2}\right)^2$ takes its least value 0 at $x=\frac{1}{2}$ and at this point $\sin 3 \pi x$ also takes its least value -1 . Hence, f(x) attains its least value 9 at $x=\frac{1}{2}$.