Question

Let M and m be respectively the absolute maximum and the absolute minimum values of the function $f\left( x \right) = 2{x^3} - 9{x^2} + 12x + 5$ in the interval $\left[ {0,3} \right]$ . Then M-m is equal to
A). 1
B). 5
C). 4
D). 9

Answer 1

Hint- In this question we have to find minimum and maximum value of the function which always occurs either at the terminal value of the function which is 0 and 3 here or at the maxima or minima domain value where the first derivative of the function gives zero .So first of all check for all values at these points.

Complete step-by-step solution -
 The given function is $f\left( x \right) = 2{x^3} - 9{x^2} + 12x + 5$
In order to find the maxima and minima of this function we will differentiate it with respect to x
$ \Rightarrow f\left( x \right) = 2{x^3} - 9{x^2} + 12x + 5 \\
   \Rightarrow f'\left( x \right) = 6{x^2} - 18x + 12 \\ $
Now the second step is to equate the first derivative equal to zero.
$ \Rightarrow f'\left( x \right) = 6{x^2} - 18x + 12 = 0$
Find the solution of the above quadratic equation
$ \Rightarrow 6{x^2} - 18x + 12 = 0 \\
   \Rightarrow {x^2} - 3x + 2 = 0 \\
   \Rightarrow \left( {x - 2} \right)\left( {x - 1} \right) = 0 \\
   \Rightarrow x = 1\;{\text{or 2}} \\ $
The numbers of values of x we get are 1, 2, 0 and 3.
We will check the value of the function at all these points and see the point where we get maximum and the minimum value.
$f\left( x \right) = 2{x^3} - 9{x^2} + 12x + 5$
Substituting x = 0 in the above equation, we get
$ f\left( 0 \right) = 2 \times {0^3} - 9 \times {0^2} + 12 \times 0 + 5 \\
  f\left( 0 \right) = 5 \\ $
For x = 1
$f\left( x \right) = 2{x^3} - 9{x^2} + 12x + 5$
Substituting x = 1 in the above equation, we get
$ f\left( 1 \right) = 2 \times {1^3} - 9 \times {1^2} + 12 \times 1 + 5 \\
  f\left( 1 \right) = 2 - 9 + 17 \\
  f\left( 1 \right) = 10 \\ $
For x = 2
$f\left( x \right) = 2{x^3} - 9{x^2} + 12x + 5$
Substituting x = 2 in the above equation, we get
$ f\left( 2 \right) = 2 \times {2^3} - 9 \times {2^2} + 12 \times 2 + 5 \\
  f\left( 2 \right) = 16 - 36 + 24 + 5 \\
  f\left( 2 \right) = 9 \\ $
For x = 3
$f\left( x \right) = 2{x^3} - 9{x^2} + 12x + 5$
Substituting x = 3 in the above equation, we get
$ f\left( 3 \right) = 2 \times {3^3} - 9 \times {3^2} + 12 \times 3 + 5 \\
  f\left( 3 \right) = 54 - 81 + 36 + 5 \\
  f\left( 3 \right) = 14 \\ $
From the above values it is clear that the function has its maximum value at x = 3 and minimum value at x = 0.
Therefore M = 14 and m = 5
The difference between the absolute maxima and the absolute minimum is
M-m = 14-5 = 9
Hence, the correct option is D.

Note- In order to solve these types of questions, you need to have a good concept of functions and the applications of derivatives. The above question can be solved by calculating the first derivative as we did in the function to get the points of maxima and minima and then the second derivative to find out which point is maxima and minima but in this case we don't calculate the second derivative and we substituted the points in the function to find the maximum and the minimum value and we see that the maxima and minima occur at the extreme points of the interval. So, in a closed interval you need to check the extreme points of a function to find the absolute maxima and minima.