Question

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

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

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

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

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

Answer 1

Given, function is $f\left(x\right)=(2x-1{)}^2+3$
It can be observed that $(2x-1{)}^2\ge 0$ for every $xϵR$
[Since, it is perfect square for any real number]
Therefore, $f\left(x\right)=(2x-1{)}^2+3\ge 3$ for every $xϵR$
The minimum value of f is attained when $2x-1=0$
i.e., $2x-1=0\Rightarrow x=\frac{1}{2}$
$\therefore$ Minimum value of $f=f\left(\frac{1}{2}\right)=(2\times \frac{1}{2}-1)+3=3$
For any value of x, value of f(x) will always be greater than 3,
hence, function f does not have particular maximum value.

Given, function is $f\left(x\right)=9x^2+12x+2=9x^2+12x+4-2$
[we add and subtract 2 for making it perfect square]
$=(9x^2+6x+6x+4)-2=\left[3x\right(3x+2)+2(3x+2\left)\right]-2=\left[\right(3x+2\left)\right(3x+2\left)\right]-2=(3x+2{)}^2-2$
It can be observed that $(3x+2{)}^2\ge 0foreveryxϵR$
Therefore, $f\left(x\right)=(3x+2{)}^2-2\ge -2foreveryxϵR$
The minimum value of f is attained when $3x+2=0$
i.e., $3x+2=0\Rightarrow x=\frac{-2}{3}$
$\therefore$ Minimum value of $f=f\left(\frac{-2}{3}\right)={(3\times \frac{-2}{3}+2)}^2-2=-2$
For any value of x, $f\left(x\right)\ge -2$, hence function f does not have a particular maximum value.

Given, function is $f\left(x\right)=-(x-1{)}^2+10$
It can be observed that $(x-1{)}^2\ge 0$ for all $xϵR$
$\Rightarrow -(x-1{)}^2\le 0$ for every $xϵR$
Therefore, $f\left(x\right)=-(x-1{)}^2+10\le 10$ for every $xϵR$
The maximum value of f is attained when (x-1)=0
i.e., $(x-1)=0\Rightarrow x=1\therefore \text{Maximum value of}f=f\left(1\right)=-(1-1{)}^2+10=10$
For any value of x, $f\left(x\right)\le 10$, hence function f does not have a particular minimum value.

Given, function is $g\left(x\right)=x^3+1$
We observe that the value of f(x) increases when the value of x increase and f(x) can be made as large as we please by giving large value to x, So, f(x) does not have a maximum value. Similarly, f(x) can be made as small as we please by giving smaller values of x. So, f(x) does not have the minimum value.

Alternate method:

Here, $g\left(x\right)=x^3+1,g^{\prime }\left(x\right)=3x^2,g"\left(x\right)=6x$
For maxima or minima put g'(x)=0
$\Rightarrow 3x^2=0\Rightarrow x=0$
At $x=0,g"\left(0\right)=6\times 0=0$
Hence at x=0, g(x) is neither maxima nor minima. It is a point of inflection.