Question

The maximum value of the function $2{x^3} - 15{x^2} + 36x + 4$ is attained at:
A) 0
B) 3
C) 4
D) 2
E) 5

Answer 1

Hint:
Here, in the question we have to find the maximum value of the function. So, first of all we will first derivate the function to obtain the critical points and then we will second derivative and put the critical points in the second derivative to obtain the maximum value in it.

Complete step by step solution:
Here, we will derive the function
 $
= 2{x^3} - 15{x^2} + 36x + 4 \\
 = 6{x^2} - 30x + 36 \\
 $
For the critical point we will take
 $
   \Rightarrow {f^{'}}(x) = 0 \\
   \Rightarrow 6{x^2} - 30x + 36 = 0 \\
   \Rightarrow 6({x^2} - 5x + 6) = 0 \\
   \Rightarrow {x^2} - 5x + 6 = 0 \\
   \Rightarrow {x^2} - 2x - 3x + 6 = 0 \\
   \Rightarrow x(x - 2) - 3(x - 3) = 0 \\
   \Rightarrow x = 2,3 \\
 $
For second derivative
 $ \Rightarrow {f^{''}}(x) = 12x - 30$
Putting the critical points in the second derivative,
 $ \Rightarrow {f^{''}}(2) = 12(2) - 30$
 $
 = 24 - 30 \\
 = - 6 < 0 \\
 $
 $ \Rightarrow {f^{''}}(3) = 12(3) - 30$
$
= 36 - 30 \\
= 6 > 0 \\
 $
From the above discussion it is clear that at point 2 we attain the maximum value.

Note:
we will also first find the second derivative and then critical points. The above method we use is known as the second derivative test. We have to substitute critical points in the second derivative not in the initial function given to us.