Question

Find the maxima and minima of the function $f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1$.

Answer 1

Hint: First find the first derivative of the function and put it to zero to find the critical points. Then find the second derivative of the function at these critical points. If the second derivative is negative then the point is of absolute maxima and if it is positive then the point is of absolute minima.

Complete step-by-step answer:
According to the question, the given function is:
$ \Rightarrow f\left( x \right) = 2{x^3} - 15{x^2} + 36x + 1$
Differentiating it with respect to $x$, we’ll get:
$ \Rightarrow f'\left( x \right) = 6{x^2} - 30x + 36$
For finding the critical points, putting first derivative to zero, we’ll get:
$
   \Rightarrow 6{x^2} - 30x + 36 = 0 \\
   \Rightarrow {x^2} - 5x + 6 = 0
 $
Factoring the quadratic equation and finding the roots, we’ll get:
$
   \Rightarrow {x^2} - 3x - 2x + 6 = 0 \\
   \Rightarrow x\left( {x - 3} \right) - 2\left( {x - 3} \right) = 0 \\
   \Rightarrow \left( {x - 2} \right)\left( {x - 3} \right) = 0 \\
   \Rightarrow x = 2{\text{ or }}x = 3
 $
Thus, $x = 2$ and $x = 3$ are two critical points of $f\left( x \right)$.
Differentiating $f'\left( x \right)$ to determine the second derivative of the function $f\left( x \right)$, we’ll get:
$
   \Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\left( {6{x^2} - 30x + 36} \right) \\
   \Rightarrow f''\left( x \right) = 12x - 30
 $
Putting $x = 2$ in $f''\left( x \right)$, we’ll get:
$
   \Rightarrow f''\left( 2 \right) = 12 \times 2 - 30 \\
   \Rightarrow f''\left( 2 \right) = 24 - 30 \\
   \Rightarrow f''\left( 2 \right) = - 6
 $
Since the second derivative is negative at $x = 2$, this is a point of absolute maxima. Now putting $x = 2$ in $f\left( x \right)$ to determine this maximum value, we’ll get:
$
   \Rightarrow {f_{\max }} = f\left( 2 \right) = 2{\left( 2 \right)^3} - 15{\left( 2 \right)^2} + 36 \times 2 + 1 \\
   \Rightarrow {f_{\max }} = f\left( 2 \right) = 16 - 60 + 72 + 1 \\
   \Rightarrow {f_{\max }} = f\left( 2 \right) = 29
 $
Now putting $x = 3$ in $f''\left( x \right)$, we’ll get:
$
   \Rightarrow f''\left( 3 \right) = 12 \times 3 - 30 \\
   \Rightarrow f''\left( 3 \right) = 36 - 30 \\
   \Rightarrow f''\left( 3 \right) = 6
 $
Since the second derivative is positive at $x = 3$, this is a point of absolute minima. Now putting $x = 3$ in $f\left( x \right)$ to determine this minimum value, we’ll get:
$
   \Rightarrow {f_{\min }} = f\left( 3 \right) = 2{\left( 3 \right)^3} - 15{\left( 3 \right)^2} + 36 \times 3 + 1 \\
   \Rightarrow {f_{\min }} = f\left( 3 \right) = 54 - 135 + 108 + 1 \\
   \Rightarrow {f_{\min }} = f\left( 3 \right) = 28
 $

Thus the point $x = 2$ is the point of absolute maxima and this maximum value of the function is 29. And the point $x = 3$ is the point of absolute minima and this minimum value of the function is 28.

Note: In case if the second derivative is also zero at any critical point of the function, then this point is neither maxima or minima. This point is called the point of inflection of the function. This is where the second derivative changes its sign from negative to positive or from positive to negative.