Question

The value of the function $\left( {x - 1} \right){\left( {x - 2} \right)^2}$ at its maxima is ;
$\left( 1 \right)1$
$\left( 2 \right)2$
$\left( 3 \right)0$
$\left( 4 \right)\dfrac{4}{{27}}$

Answer 1

There are certain steps which need to be followed to solve this type of problems: $\left( 1 \right){\text{Let }}y = f\left( x \right)$ $\left( 2 \right){\text{Differentiate with respect to }}x{\text{ to find out }}\dfrac{{dy}}{{dx}}$ . $\left( 3 \right)$ To find out maxima or minima of a given function equate $\dfrac{{dy}}{{dx}} = 0$ . $\left( 4 \right)$ We will get all the possible values of $x$ depending on the degree of the given equation. $\left( 5 \right)$ Find $\dfrac{{{d^2}y}}{{d{x^2}}}$ . $\left( 6 \right)$ Put the values of $x$ one by one in $\dfrac{{{d^2}y}}{{d{x^2}}}$ $\left( 7 \right)$ If $\dfrac{{{d^2}y}}{{d{x^2}}} \succ 0$ i.e. if $\dfrac{{{d^2}y}}{{d{x^2}}}$ is positive then this is called point of minima and that value of $y$ is minimum for that particular value of $x$ . $\left( 8 \right)$ If $\dfrac{{{d^2}y}}{{d{x^2}}} \prec 0$ i.e. if $\dfrac{{{d^2}y}}{{d{x^2}}}$ is negative then this is called point of maxima and that value of $y$ is maximum for that particular value of $x$.

Complete step by step solution:
The function given is ;
$\Rightarrow f\left( x \right) = \left( {x - 1} \right){\left( {x - 2} \right)^2}$
Let us simplify the above expression, so that it will be easier to differentiate;
$\left( {\because {{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right)$
Applying the above formula in the given expression, we get;
$\Rightarrow f\left( x \right) = \left( {x - 1} \right)\left( {{x^2} - 4x + 4} \right)$
On multiplication;
$\Rightarrow f\left( x \right) = {x^3} - 4{x^2} + 4x - {x^2} + 4x - 4$
On further simplification;
$\Rightarrow f\left( x \right) = {x^3} - 5{x^2} + 8x - 4$
Differentiating the above expression with respect to $x,$ we get;
$\Rightarrow f'\left( x \right) = 3{x^2} - 10x + 8{\text{ }}......\left( 1 \right)$
Factorizing the above equation, to get the values of $x$ ;
$\Rightarrow f'\left( x \right) = 3{x^2} - 6x - 4x + 8$
$\Rightarrow f'\left( x \right) = 3x\left( {x - 2} \right) - 4\left( {x - 2} \right)$
$\Rightarrow f'\left( x \right) = \left( {3x - 4} \right)\left( {x - 2} \right)$
Now equating individual terms to zero to get the value of $x$ ,
$\Rightarrow \left( {3x - 4} \right)\left( {x - 2} \right) = 0$
We get two values of $x$ as;
$\because 3x - 4 = 0$
$\therefore x = \dfrac{4}{3}$
Now, equating the second term to zero ;
$\because x - 2 = 0$
$\therefore x = 2$
Now, let us calculate $f''\left( x \right)$ ;
$\Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}f'\left( x \right)$
$\Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\left( {3{x^2} - 10x + 8} \right)$
$\Rightarrow f''\left( x \right) = 6x - 10{\text{ }}......\left( 2 \right)$
Now, put the values of $x = \dfrac{4}{3}{\text{ and }}x = 2{\text{ in equation }}\left( 2 \right)$ , we get ;
$\Rightarrow f''\left( {\dfrac{4}{3}} \right) = 6\left( {\dfrac{4}{3}} \right) - 10 = - 2$
$\Rightarrow f''\left( {\dfrac{4}{3}} \right) = - 2{\text{ means }}f''\left( {\dfrac{4}{3}} \right) \prec 0$
Therefore, this is the point of maxima.
$\Rightarrow f''\left( 2 \right) = 6\left( 2 \right) - 10 = 2$
$\Rightarrow f''\left( 2 \right) = 2{\text{ means }}f''\left( 2 \right) \succ 0$
Therefore, this is the point of minima.
According to the question, we have to calculate point of maxima ;
$\because f\left( x \right) = \left( {x - 1} \right){\left( {x - 2} \right)^2}$
And it is maximum at $x = \dfrac{4}{3}$ , therefore put the value of $x$ in the above equation ;
$\therefore f\left( {\dfrac{4}{3}} \right) = \left( {\dfrac{4}{3} - 1} \right){\left( {\dfrac{4}{3} - 2} \right)^2}$
Simplifying the above expression;
$\Rightarrow f\left( {\dfrac{4}{3}} \right) = \left( {\dfrac{1}{3}} \right) \times {\left( {\dfrac{{ - 2}}{3}} \right)^2}$
$\Rightarrow f\left( {\dfrac{4}{3}} \right) = \dfrac{4}{{27}}$
Therefore, the value of the function $\left( {x - 1} \right){\left( {x - 2} \right)^2}$ at its maxima i.e. $\dfrac{4}{3}$ is $\dfrac{4}{{27}}$ .
Hence the correct answer for this question is option $\left( 4 \right)$ i.e., $\dfrac{4}{{27}}$.

Note:
The point of maxima and minima for a given function can also be termed as the extreme values for that particular function. Maxima means the highest value of the function, there is the concept of global maxima and local maxima. Global maxima means the highest value of the function, the function can not exceed this value, while the local maxima means there can be some other values of the function in the domain for which it’s value can be higher than the local maxima. We have used the second derivative test in this question which is used to determine whether a given point is point of maxima or point of minima. $\left( 1 \right)$ If ${\left. {\dfrac{{{d^2}y}}{{d{x^2}}}} \right|_{x = {x_0}}} \prec 0$ means rate of change is negative then $x = {x_0}$ is called point of maxima . $\left( 2 \right)$ ${\left. {\dfrac{{{d^2}y}}{{d{x^2}}}} \right|_{x = {x_0}}} \succ 0$ means the rate of change is negative then $x = {x_0}$ is called the point of minima.