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

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

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

(iii) $f\left( x \right) = - {\left( {x - {\rm{1}}} \right)^2} + 10$

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

(i) We have, $f\left( x \right) = {\left( {2x - 1} \right)^2}{\rm{ + }}3$ for all $x \in R$.

Since ${\left( {2x - 1} \right)^2} \ge 0 \Rightarrow {\left( {2x - 1} \right)^2} + 3 \ge 3$

Therefore the minimum value of $f\left( x \right) = 3$ , which occurs when $2x - 1 = 0$ i.e., when $x = 1/2$

Value of $f\left( x \right)$ has no maximum value because $f\left( x \right) \to \infty \,\,{\rm{as}}\,\,\left| x \right| \to \infty$

(ii) We have $f(x) = 9{x^2} + 12x + 2 = 9\left( {{x^2} + \cfrac{4}{3}x} \right) + 2$
$= 9\left\{ {{x^2} + \cfrac{4}{3}x + \cfrac{4}{9}} \right\} + 2 - 4 = 9{\left( {x + \cfrac{2}{3}} \right)^2} - 2$

Since ${\left( {x + \cfrac{2}{3}} \right)^2} \ge 0 = 9{\left( {x + \cfrac{2}{3}} \right)^2} - 2 \ge - 2$

$\Rightarrow f\langle x) \ge - 2$ for all $x \in R$.

Therefore the minimum, $f(x) = - 2$, which occurs when $x + \cfrac{2}{3} = 0$, i.e., when $x = \cfrac{{ - 2}}{3}$.
$f(x)$ has no maximum value because $f(x\mathop )\limits^` \to \infty$ as $|x| \to \infty$.

(iii) We have, $f(x) = 10 - {(x - 1)^2}$ for aI $x \in R$.
${(x - 1)^2} \ge 0\,\,\forall \,\,x \in R \Rightarrow - {(x - 1)^2} \le 0\,\,\forall \,\,x \in R$

$\Rightarrow 10 - {(x - 1)^2} \le 10\,\,\forall \,\,x \in R$

Therefore the maximum $f(x) = 10$, which occurs when $x - 1 = 0$ i.e., when $x = 1$.

$f(x)$ has no minimum value because $f(x) \to - \infty$ as $|x| \to \infty$.

(iv) We have, $g(x) = {x^3} + 1$
As $x \to \infty ,g(x) \to \infty$ and as $x \to - \infty ,g(x) \to - \infty$
hence we can say that $g(x)$ has neither a maximum nor a minimum value.