The largest value of $2{x^3} - 3{x^2} - 12x + 5$ for $ - 2 \leqslant x \leqslant 4$ occurs at $x$ is equal to
A. $ - 4$
B. $0$
C. $1$
D. $4$

Hint: To find the maximum and minimum point, differentiate the given function $f(x) = 2{x^3} - 3{x^2} - 12x + 5$ with respect to $x$ the put the required derivative with equal to zero and find the roots. Again, differentiate the function and put the required values of $x$ in the double derivative function. Lastly, to check, put the values of $x$ in the given function.

Formula Used:
Differentiation formula –
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Complete step by step solution:
Given that,
$f(x) = 2{x^3} - 3{x^2} - 12x + 5 - - - - - (1)$
Differentiate equation (1) with respect to $x$
$f'(x) = 6{x^2} - 6x - 12 - - - - - (2)$
Let, equation (2) be equal to zero
$f'(x) = 6{x^2} - 6x - 12 = 0$
$ \Rightarrow 6{x^2} - 6x - 12 = 0$
${x^2} - x - 2 = 0$
${x^2} - 2x + x - 2 = 0$
$x(x - 2) + 1(x - 2) = 0$
$(x + 1)(x - 2) = 0$
$x + 1 = 0,x - 2 = 0$
$x = - 1,x = 2$
Differentiate equation (2) with respect to $x$
$f''(x) = 12x - 6$
At $x = - 1$
$f''( - 1) = 12( - 1) - 6$
$f''( - 1) = - 18 < 0$
$ \Rightarrow x = - 1$ can be a point of maximum
At $x = 2$
$f''(2) = 12(2) - 6$
$f''(2) = 18 > 0$
$ \Rightarrow x = 2$ is the point of minimum value
Now, to check the values of $f(x)$ put $x = - 2,2, - 1,4$($ - 2 \leqslant x \leqslant 4$ given)
At $x = - 2$,
$f( - 2) = 2{\left( { - 2} \right)^3} - 3{\left( { - 2} \right)^2} - 12\left( { - 2} \right) + 5 = 1$
At $x = 2$,
$f(2) = 2{\left( 2 \right)^3} - 3{\left( 2 \right)^2} - 12\left( 2 \right) + 5 = - 15$
At $x = - 1$,
$f( - 1) = 2{\left( { - 1} \right)^3} - 3{\left( { - 1} \right)^2} - 12\left( { - 1} \right) + 5 = 12$
At $x = 4$,
$f(4) = 2{\left( 4 \right)^3} - 3{\left( 4 \right)^2} - 12\left( 4 \right) + 5 = 37$
Hence, a t $x = 4$the value of $f(x)$ is maximum.

Option ‘D’ is correct

Note: (General quadratic equation $a{x^2} + bx + c$)In such questions, while finding the roots of quadratic equation multiply the value of $a,c$ and then try to open that multiple in such form where you will get the value of $b$. For example – Let, the quadratic equation be $2{x^2} + 3x - 2$. Here, the multiple of $a,c$ is $ - 4$ and it can be written as $ + 4 - 1 = 3 = b$.