Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions: $f(x) = 4x^{2 }- 4x + 4$ on $R.$
✓
Answer
$\text{f}(\text{x})=4\text{x}^{2}-4\text{x}+4$ on $R$
$=4\text{x}^{2}-4\text{x}+1+3$
$=(2\text{x})^{2}+3$
$ \therefore(2\text{x}-1)^{2}\geq0$
$\Rightarrow(2\text{x}-1)^{2}+3\geq3$
$\Rightarrow\text{f}(\text{x})\geq\text{f}\Big(\frac{1}{2}\Big)$
Thus, the minimum value of $f(x)$ is $3$ at $\text{x}=\frac{1}{2}$
since, $f(x)$ can be made as large as. Therefore maximum value does not exist.
$=4\text{x}^{2}-4\text{x}+1+3$
$=(2\text{x})^{2}+3$
$ \therefore(2\text{x}-1)^{2}\geq0$
$\Rightarrow(2\text{x}-1)^{2}+3\geq3$
$\Rightarrow\text{f}(\text{x})\geq\text{f}\Big(\frac{1}{2}\Big)$
Thus, the minimum value of $f(x)$ is $3$ at $\text{x}=\frac{1}{2}$
since, $f(x)$ can be made as large as. Therefore maximum value does not exist.
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