Question

How do you find the exact relative maximum and minimum of the polynomial function of $f\left( x \right) = 2{x^3} - 3{x^2} - 12x$ ?

Answer 1

in this type of problem, we use the concept of derivatives of the polynomial. To find the relative maximum or minimum of a given polynomial, first we differentiate the given polynomial with respect to the variable and then we equate it equal to the zero. When we equate the derivative of the given polynomial then we will find all points at which the given polynomial is maximum or minimum. The given polynomial will have maxima at the point on which the polynomial will give maximum value. In the same manner the point at which the value of the given polynomial will be minimum, the respective point will be the minima of the polynomial. To find the value of the given polynomial, we will substitute the point in the equation of the polynomial and find the value of the function.

Complete step by step solution:
Step: 1 the given polynomial is,
$f\left( x \right) = 2{x^3} - 3{x^2} - 12x$
To find the points at which the polynomial is maximum or minimum, differentiate the given equation of the polynomial with respect to $x$.
Let
$f\left( x \right) = y$
Therefore,
$\Rightarrow y = 2{x^3} - 3{x^2} - 12x$
Step: 2 differentiate the equation with respect to $x$ .
$\Rightarrow \dfrac{{dy}}{{dx}} = 2\dfrac{d}{{dx}}{x^3} - 3\dfrac{d}{{dx}}{x^2} - 12\dfrac{d}{{dx}}x \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 2 \times 3{x^2} - 3 \times 2x - 12 \\\$
Simplify the equation and equate it with equal to zero.
$\Rightarrow 6{x^2} - 6x - 12 = 0$
Solve the given quadratic equation to find the value of $x$.
Divide the both sides of the equation with $6$.
$\Rightarrow \dfrac{{6{x^2} - 6x - 12}}{6} = \dfrac{0}{6} \\\ \Rightarrow {x^2} - x - 2 = 0 \\\$
Take constant terms in the equation on the right hand side of the equation.
$\Rightarrow {x^2} - x - 2 = 0 \\\ \Rightarrow {x^2} - x = 2 \\\$
Step: 3 solve the quadratic equation by completing the square.
The coefficient of ${x^2}$ is one. So no need to divide with one. The coefficient of $x$ is $\left( { - 1} \right)$ and half of it is $\left( { - \dfrac{1}{2}} \right)$. Square the half and add both sides of the equation.
Square of the half is $\left( {\dfrac{1}{4}} \right)$ so add it to both sides of the equation.
$\Rightarrow {x^2} - x + \dfrac{1}{4} = 2 + \dfrac{1}{4}$
Now make the perfect square of the equation by using the formula,
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
Therefore,
$\Rightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{9}{4}$
Take the square root to both sides of the equation.
$\Rightarrow \sqrt {{{\left( {x - \dfrac{1}{2}} \right)}^2}} = = \sqrt {\dfrac{9}{4}} \\\ \Rightarrow x - \dfrac{1}{2} = \pm \dfrac{3}{2} \\\$
Consider the positive value of $x$ and solve for the $x$.
$\Rightarrow - \dfrac{1}{2} = + \dfrac{3}{2} \\\ \Rightarrow x = \dfrac{3}{2} + \dfrac{1}{2} \\\ \Rightarrow x = 2 \\\$
Consider the negative value of $x$.
$\Rightarrow x - \dfrac{1}{2} = - \dfrac{3}{2} \\\ \Rightarrow x = - \dfrac{3}{2} + \dfrac{1}{2} \\\ \Rightarrow x = - 1 \\\$
So the points are $x = - 1,2$
Step: 4 to find the maximum or minimum value of the given function, substitute $x = - 1$ in the polynomial.
$\Rightarrow f\left( x \right) = 2{x^3} - 3{x^2} - 12x \\\ \Rightarrow f\left( { - 1} \right) = 2{\left( { - 1} \right)^3} - 3{\left( { - 1} \right)^2} - 12\left( { - 1} \right) \\\ \Rightarrow f\left( { - 1} \right) = - 2 - 3 + 12 \\\ \Rightarrow f\left( { - 1} \right) = 7 \\\$
Now substitute $x = 2$ in the equation of polynomial and solve.
$\Rightarrow f\left( 2 \right) = 2{\left( 2 \right)^3} - 3{\left( 2 \right)^2} - 12\left( 2 \right) \\\ \Rightarrow f\left( 2 \right) = 16 - 12 - 24 \\\ \Rightarrow f\left( 2 \right) = - 20 \\\$
**
Therefore the maximum value of the polynomial is 7 and minimum value of the polynomial is -20.**

Note:
Students must know to solve the quadratic equation by using the complete square method. They must know to differentiate the given function with respect to the given variable. Students are advised to not make any mistakes while solving the equations. They should calculate the maximum or minimum value of the polynomial by substituting the given point in the equation.