Question

The maximum value of $\left( {{\text{x}} - 1} \right)\left( {{\text{x}} - 2} \right)\left( {{\text{x}} - 3} \right)$ is

Answer 1

Hint: To find the maximum value, we need to differentiate the function and equate it to zero, then again differentiating the function will give us maxima if its value is less than zero. We will apply product rule here which is given by-
${\text{f}}\left( {\text{x}} \right) = {\text{u}}.{\text{v}}$
$\dfrac{{d{\text{f}}\left( {\text{x}} \right)}}{{d{\text{x}}}} = {\text{u}}\dfrac{{d{\text{v}}}}{{d{\text{x}}}} + {\text{v}}\dfrac{{d{\text{u}}}}{{d{\text{x}}}}$

Complete step by step solution:
Let f(x) = (x - 1)(x - 2)(x - 3)
We will differentiate f(x) with respect to x and equate it to zero. This will give us the critical points of the function.
f’(x) = (x - 2)(x - 3) + (x - 1)(x - 3) +(x - 1)(x - 2) = 0
f’(x) = ${{x}^{2}}$ -5x + 6 + ${{x}^{2}}$ -4x + 3 + ${{x}^{2}}$ -3x + 2 = 0
f’(x) = 3${{x}^{2}}$ -12x +11 = 0
Using the quadratic formula-
$=\dfrac{-\mathrm b\pm\sqrt{\mathrm b^2-4\mathrm{ac}}}{2\mathrm a}$
$=\dfrac{12\pm\sqrt{144-4\times3\times11}}{2\times3}$
$=\dfrac{12\pm\sqrt{12}}6=2\pm\dfrac1{\sqrt3}$
These are the two roots. Differentiating f’(x) with respect to x again,
f’’(x) = 6x - 12,
This function is the double differential of the given function. On substituting the critical points here, if the value of the function is negative, the critical point will be a point of maxima. On the other hand, if the value of the comes out to be positive, then the point substituted is a minima.
Substituting the roots in f’’(x),
$\mathrm f''\left(2+\dfrac1{\sqrt3}\right)=12+\dfrac6{\sqrt3}-12=\dfrac6{\sqrt3}>0\\\mathrm f''\left(2-\dfrac1{\sqrt3}\right)=12-\dfrac6{\sqrt3}-12=\dfrac{-6}{\sqrt3}<0$
Hence, the second root gives the maxima condition because the function has a negative value. To find the maximum value of the original function, we will substitute the this critical point in the function as-
$f\left(2-\dfrac1{\sqrt3}\right)=\left(2-\dfrac1{\sqrt3}-1\right)\left(2-\dfrac1{\sqrt3}-2\right)\left(2-\dfrac1{\sqrt3}-3\right)\\f\left(2-\dfrac1{\sqrt3}\right)=\left(1-\dfrac1{\sqrt3}\right)\left(1+\dfrac1{\sqrt3}\right)\left(\dfrac1{\sqrt3}\right)\\f\left(2-\dfrac1{\sqrt3}\right)=\dfrac2{3\sqrt3}$

Note: In this question, multiple concepts are used including quadratic equations, differentiation and concept of maxima and minima. All these need to be thoroughly revised before solving such types of questions. A common mistake here is that students reverse the maxima and minima conditions, hence they end up finding the minima instead of maxima or vice versa.