Question:
$f(x)=4 x^{2}-4 x+4$ on $R$
Solution:
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.