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Given: $f(x)=4 x^{2}-4 x+4$

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

$\Rightarrow f(x)=(2 x-1)^{2}+3$

Now,

$(2 x-1)^{2} \geq 0$ for all $x \in R$

$\Rightarrow f(x)=(2 x-1)^{2}+3 \geq 3$ for all $x \in R$

$\Rightarrow f(x) \geq 3$ for all $x \in R$

The minimum value of $f$ is attained when $(x-1)=0$.

$(2 x-1)=0$

$\Rightarrow x=\frac{1}{2}$

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.