Find the maximum and minimum value, f(x) = (2x - 1)^2 + 3 - Vidyadip

Question

Find the maximum and minimum value, $f(x) = (2x - 1)^2 + 3$

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Answer

It is given that $f(x)=(2 x-1)^2+3$
Now, we can see that $(2 x-1)^2 \geq 0$ for every $x \in R$
$\Rightarrow f(x)=(2 x-1)^2+3 \geq 3 \text { for every } x \in R$
The minimum value of $f$ is attained when $2 x-1=0$
$ 2 x-1=0 $
$ \Rightarrow x=\frac{1}{2}$
Then, Minimum value of $f=f\left(\frac{1}{2}\right)=\left(2 \cdot \frac{1}{2}-1\right)^2+3=3$
Now, $\mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}-2=0, \Rightarrow \mathrm{x}=\frac{1}{2}$ is the only critical point which is a minimum.
Therefore, function $f$ does not have a maximum value.

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