# Find the maximum and minimum values, if any,

Question:

Find the maximum and minimum values, if any, of the following functions given by

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

(ii) $f(x)=9 x^{2}+12 x+2$

(iii) $f(x)=-(x-1)^{2}+10$

(iv) $g(x)=x^{3}+1$

Solution:

(i) The given function is $f(x)=(2 x-1)^{2}+3$.

It can be observed that $(2 x-1)^{2} \geq 0$ for every $x \in \mathbf{R}$.

Therefore, $f(x)=(2 x-1)^{2}+3 \geq 3$ for every $x \in \mathbf{R}$.

The minimum value of $f$ is attained when $2 x-1=0$.

$2 x-1=0 \Rightarrow x=\frac{1}{2}$

$\therefore$ Minimum value of $f=f\left(\frac{1}{2}\right)=\left(2 \cdot \frac{1}{2}-1\right)^{2}+3=3$

Hence, function f does not have a maximum value.

(ii) The given function is $f(x)=9 x^{2}+12 x+2=(3 x+2)^{2}-2$.

It can be observed that $(3 x+2)^{2} \geq 0$ for every $x \in \mathbf{R}$.

Therefore, $f(x)=(3 x+2)^{2}-2 \geq-2$ for every $x \in \mathbf{R}$.

The minimum value of $f$ is attained when $3 x+2=0$.

$3 x+2=0 \Rightarrow x=\frac{-2}{3}$

$\therefore$ Minimum value of $f=f\left(-\frac{2}{3}\right)=\left(3\left(\frac{-2}{3}\right)+2\right)^{2}-2=-2$

Hence, function f does not have a maximum value.

(iii) The given function is $f(x)=-(x-1)^{2}+10$.

It can be observed that $(x-1)^{2} \geq 0$ for every $x \in \mathbf{R}$.

Therefore, $f(x)=-(x-1)^{2}+10 \leq 10$ for every $x \in \mathbf{R}$.

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

$(x-1)=0 \Rightarrow x=1$

$\therefore$ Maximum value of $f=f(1)=-(1-1)^{2}+10=10$

Hence, function $f$ does not have a minimum value.

(iv) The given function is $g(x)=x^{3}+1$.

Hence, function $g$ neither has a maximum value nor a minimum value.