Question

How do you find the exact relative maximum and minimum of the polynomial function of $f\left( x \right) = {x^3} - 6{x^2} + 15$?

Answer 1

Hint: First, we should find the differentiation of function \[f\left( x \right)\]. Let us assume the differentiation of function f(x) is equal to $f'\left( x \right)$ . Now we should find the value of x where the function $f'\left( x \right)$ is equal to zero. We know that a function \[f\left( x \right)\] will have a relative maximum at \[x = a\] where $f'\left( a \right) = 0$ and $f''\left( a \right) < 0$. We also know that a function \[f\left( x \right)\] will have a relative minimum at \[x = a\] where $f'\left( a \right) = 0$ and $f''\left( a \right) > 0$. So, to know the relative maximum and relative minima of a function \[f\left( x \right)\] we have to find the $f''\left( x \right)$. Now we have to substitute the value of $x$ where the $f'\left( x \right)$ is equal to zero. This will give the relative maxima and relative minima of the function \[f\left( x \right)\]. Now by substituting the respective values of $x$, we can get the respective relative maximum and relative minimum.

Complete step-by-step answer:
Before solving the question, we should have an idea of relative maxima and relative minima. By using this relative minimum and the relative maximum we can find the values of relative maximum and relative minimum.
From the question, we were given that $f\left( x \right) = {x^3} - 6{x^2} + 15$.
A function \[f\left( x \right)\] will have relative maxima and relative minima at the value of \[x\] where \[f'\left( x \right)\] is equal to zero.
$f'\left( x \right) = \dfrac{d}{{dx}}f\left( x \right)$
Substitute the value of the function,
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^3} - 6{x^2} + 15} \right)$
Open the bracket and break them into parts,
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^3}} \right) - \dfrac{d}{{dx}}\left( {6{x^2}} \right) + \dfrac{d}{{dx}}\left( {15} \right)$
Differentiate the terms,
$ \Rightarrow f'\left( x \right) = 3{x^2} - 12x$
Now we have to find the value of x where $f'\left( x \right)$ is equal to zero.
$ \Rightarrow f'\left( x \right) = 0$
Substitute the values,
$ \Rightarrow 3{x^2} - 12x = 0$
Take common from the terms,
$ \Rightarrow 3x\left( {x - 4} \right) = 0$
So, the value of $x$ will be,
$ \Rightarrow x = 0,4$
At $x = 0,4$ , we will have the value of $f'\left( x \right)$ equal to zero.
So, we will have a relative maximum or relative minimum at $x = 0,4$ for $f\left( x \right) = {x^3} - 6{x^2} + 15$.
We know that a function \[f\left( x \right)\] will have a relative maximum at \[x = a\] where $f'\left( a \right) = 0$ and $f''\left( a \right) < 0$.
We also know that a function \[f\left( x \right)\] will have a relative minimum at \[x = a\] where $f'\left( a \right) = 0$ and $f''\left( a \right) < 0$.
Now find the double derivative of $f\left( x \right)$.
$f''\left( x \right) = \dfrac{{{d^2}}}{{d{x^2}}}f\left( x \right)$
As we know, $f'\left( x \right) = \dfrac{d}{{dx}}f\left( x \right)$. Then,
$ \Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}f'\left( x \right)$
Substitute the value,
$ \Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\left( {3{x^2} - 12x} \right)$
Open the bracket and break them into parts,
$ \Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\left( {3{x^2}} \right) - \dfrac{d}{{dx}}\left( {12x} \right)$
Differentiate the terms,
$ \Rightarrow f''\left( x \right) = 6x - 12$ ….. (1)
To check whether $x = 0$ is a relative maximum or minimum, we should find the value of $f''\left( 0 \right)$.
Now we will substitute \[x = 0\] in equation (1).
$ \Rightarrow f''\left( 0 \right) = 6\left( 0 \right) - 12$
Simplify the terms,
$ \Rightarrow f''\left( 0 \right) = - 12$
The value of $f''\left( 0 \right)$ is equal to -12. It is clear that $f''\left( 0 \right) < 0$. So, we can say that \[x = 0\] is a point of maxima.
Now we will substitute \[x = 0\] in \[f\left( x \right)\].
$ \Rightarrow f\left( 0 \right) = {\left( 0 \right)^3} - 6{\left( 0 \right)^2} + 15$
Simplify the terms,
$ \Rightarrow f\left( 0 \right) = 15$
So, the maximum value of \[f\left( x \right)\] is equal to 15.
To check whether $x = 4$ is a relative maximum or minimum, we should find the value of $f''\left( 4 \right)$.
Now we will substitute \[x = 4\] in equation (1).
$ \Rightarrow f''\left( 4 \right) = 6\left( 4 \right) - 12$
Simplify the terms,
$ \Rightarrow f''\left( 4 \right) = 24 - 12$
Subtract the values,
$ \Rightarrow f''\left( 4 \right) = 12$
The value of $f''\left( 4 \right)$ is equal to 12. It is clear that $f''\left( 4 \right) > 0$. So, we can say that \[x = 4\] is a point of minima.
Now we will substitute \[x = 4\] in \[f\left( x \right)\].
$ \Rightarrow f\left( 4 \right) = {\left( 4 \right)^3} - 6{\left( 4 \right)^2} + 15$
Simplify the terms,
$ \Rightarrow f\left( 4 \right) = 64 - 96 + 15$
Again, simplify the terms,
$ \Rightarrow f\left( 4 \right) = - 17$
So, the minimum value of \[f\left( x \right)\] is equal to -17.

Hence, the exact relative maximum and minimum of the polynomial are 15 and -17.

Note:
1) We say that a function \[f\left( x \right)\] has a relative maximum value at \[x = a\], if \[f\left( a \right)\] is greater than any value immediately preceding or following.
2) We call it a "relative" maximum because other values of the function may in fact be greater.
3) We say that a function \[f\left( x \right)\] has a relative minimum value at \[x = b\], if \[f\left( b \right)\] is less than any value immediately preceding or following.
4) The value of the function, the value of y, at either a maximum or a minimum is called an extreme value.