Question

The number of critical points of the function$f\left( x \right) = \left| {x - 1} \right|\left| {x - 2} \right|$is
A.1
B.2
C.3
D.None of these

Answer 1

First we define the function in interval $x < 1$, $1 \leqslant x < 2$and $x \geqslant 2$after that differentiate the function with respect to x in given interval. If the value of x lies in the interval then value of x is our critical point if not then it will not be a critical point. After that we calculate the left and right hand derivative of a function at branch point if the derivative does not exist at branch point then the point is also known as critical point.

Complete step-by-step answer:
We have given $f\left( x \right) = \left| {x - 1} \right|\left| {x - 2} \right|$
Define f\left( x \right) = \left\\{ \left( {x - 1} \right)\left( {x - 2} \right);x < 1 \\\ \- \left( {x - 1} \right)\left( {x - 2} \right);1 \leqslant x < 2 \\\ \left( {x - 1} \right)\left( {x - 2} \right);x \geqslant 2 \\\ \right.
\Rightarrow $$$$f\left( x \right) = \left\\{ {x^2} - 3x + 2;x < 1 \\\ \- \left( {{x^2} - 3x + 2} \right);1 \leqslant x < 2 \\\ {x^2} - 3x + 2;x \geqslant 2 \\\ \right.
For $x < 1$we have,
$f\left( x \right) = {x^2} - 3x + 2$
Differentiating with respect to x we get,
$f'\left( x \right) = 2x - 3$
As we know to find critical points we have to equate derivatives of $f\left( x \right)$ with zero.
So, $f'\left( x \right) = 0$
$\Rightarrow 2x - 3 = 0$
$\Rightarrow x = \dfrac{3}{2}$
As $x = \dfrac{3}{2} = 1.5$does not lies in $x < 1$
so $x = \dfrac{3}{2}$ is not a critical point of f(x) for $x < 1$.
Which implies that when$x < 1$the function $f\left( x \right)$does not have any critical point.
For $1 \leqslant x < 2$we have,
$f\left( x \right) = - ({x^2} - 3x + 2)$
Differentiating with respect to x we get,
$f'\left( x \right) = - \left( {2x - 3} \right)$
$\Rightarrow f'\left( x \right) = - 2x + 3$
As we know to find critical points we have to equate derivatives of $f\left( x \right)$ with zero.
So,$f'\left( x \right) = 0$
$\Rightarrow - 2x + 3 = 0$
$\Rightarrow 2x = 3$
$\Rightarrow x = \dfrac{3}{2}$
As $x = \dfrac{3}{2}$lies between$1 \leqslant x < 2$.
So, $x = \dfrac{3}{2}$is a critical point of f(x) in the interval $1 \leqslant x < 2$.
For $x \geqslant 2$we have,
$f\left( x \right) = {x^2} - 3x + 2$
Differentiating with respect to x we get,
$f'\left( x \right) = 2x - 3$
As we know to find critical points we have to equate derivatives of $f\left( x \right)$ with zero.
So, $f'\left( x \right) = 0$
$\Rightarrow 2x - 3 = 0$
$\Rightarrow x = \dfrac{3}{2}$=1.5
As $1.5 < 2$it will not lie in the interval $x \geqslant 2$
So, $x = \dfrac{3}{2}$ is not a critical point of f(x) for$x \geqslant 2$.
Which implies that when $x \geqslant 2$ the function $f\left( x \right)$does not have any critical point.
Now, we calculate the left and right hand derivative of a function at branch point if the derivative does not exist then the point is known as critical point.

2x - 3;x < 1 \\\ \- \left( {2x - 3} \right);1 \leqslant x < 2 \\\ 2x - 3;x \geqslant 2 \\\ \right.$$ For $$x = 1$$ Consider, $$Lf'\left( x \right) = $$$$2x - 3$$ $$ \ldots \left( 1 \right)$$ Now, put x=1 in equation (1) $$ \Rightarrow Lf'\left( 1 \right) = 2\left( 1 \right) - 3$$ $$ \Rightarrow Lf'\left( 1 \right) = - 1$$ $$Rf'\left( x \right) = - \left( {2x - 3} \right)$$ $$ \ldots \left( 2 \right)$$ Now, put x=1 in equation (2) $$ \Rightarrow Rf'\left( 1 \right) = - \left( {2\left( 1 \right) - 3} \right)$$ $$ \Rightarrow Rf'\left( 1 \right) = - \left( {2 - 3} \right)$$ $$ \Rightarrow Rf'\left( 1 \right) = - \left( { - 1} \right)$$ $$ \Rightarrow Rf'\left( 1 \right) = 1$$ $$ \Rightarrow Lf'\left( 1 \right) \ne Rf'\left( 1 \right)$$ $$ \Rightarrow $$$$f'\left( 1 \right)$$ does not exist. Similarly, Now we calculate for$$x = 2$$ Consider, $$Lf'\left( x \right) = - \left( {2x - 3} \right)$$ $$ \ldots \left( 3 \right)$$ Put $$x = 2$$in equation (3) $$ \Rightarrow Lf'\left( 2 \right) = - \left( {2\left( 2 \right) - 3} \right)$$ $$ \Rightarrow Lf'\left( 2 \right) = - \left( {4 - 3} \right)$$ $$ \Rightarrow Lf'\left( 2 \right) = - 1$$ $$Rf'\left( x \right) = 2x - 3$$ $$ \ldots \left( 4 \right)$$ Put $$x = 2$$in equation (4) $$ \Rightarrow Rf'\left( 2 \right) = 2\left( 2 \right) - 3$$ $$ \Rightarrow Rf'\left( x \right) = 4 - 3$$ $$ \Rightarrow Rf'\left( x \right) = 1$$ $$ \Rightarrow Lf'\left( 2 \right) \ne Rf'\left( 2 \right)$$ $$ \Rightarrow $$$$f'\left( 2 \right)$$ does not exist. Therefore, the total number of critical points is 3. Critical points of $$f\left( x \right) = \left| {x - 1} \right|\left| {x - 2} \right|$$ are $$\left\\{ {1,2,\dfrac{3}{2}} \right\\}$$ Hence, option C. 3 is the correct answer. **Note:** Critical point of a function: ‘When dealing with function of a real variable a critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero.’