Question

The function \[f\] defined by \[f\left( x \right) = {x^3} - 6{x^2} - 36x + 7\] is increasing, if
A. x < 2 and also x > 6
B. x > 2 and also x < 6
C. x < –2 and also x < 6
D. x < –2 and also x > 6

Answer 1

Hint: In this question, we need to find the condition of x for which the function \[f\] defined by \[f\left( x \right) = {x^3} - 6{x^2} - 36x + 7\] is increasing. For this, we have to find the derivative of \[f\left( x \right) = {x^3} - 6{x^2} - 36x + 7\] with respect to the variable x. Here, we will use the concept such as the function is increasing if it derivative is greater than zero.

Formula used:
The following formula of derivative is used to solve the given question.
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Also, we can say that the derivative of any constant term is always is zero.

Complete step-by-step solution:
We know that \[f\left( x \right) = {x^3} - 6{x^2} - 36x + 7\]
Let us find the condition for which the function f is increasing.
So, we can say that the function \[f\] defined by \[f\left( x \right) = {x^3} - 6{x^2} - 36x + 7\] is increasing if \[f'\left( x \right) > 0\]
Thus, we will find the derivative of \[f\left( x \right) = {x^3} - 6{x^2} - 36x + 7\] .
\[f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^3} - 6{x^2} - 36x + 7} \right)\]
By separating the terms, we get
\[f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^3}} \right) - 6\dfrac{d}{{dx}}\left( {{x^2}} \right) - 36\dfrac{d}{{dx}}\left( x \right) + 7\dfrac{d}{{dx}}\left( 1 \right)\]
But we know that \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] and the derivative of constant term is zero.
Thus, we get
\[f'\left( x \right) = 3{x^2} - 6\left( {2x} \right) - 36\left( 1 \right) + 7\left( 0 \right)\]
By simplifying, we get
\[f'\left( x \right) = 3{x^2} - 12x - 36 + 0\]
Now the function \[f\left( x \right)\] is increasing if \[f'\left( x \right) > 0\]
So, we get
\[\left( {3{x^2} - 12x - 36} \right) > 0\]
By taking 3 common from it, we get
\[3\left( {{x^2} - 4x - 12} \right) > 0\]
\[\left( {{x^2} - 4x - 12} \right) > 0\]
By factorising, we get
\[\left( {{x^2} - 6x + 2x - 12} \right) > 0\]
\[\left( {x\left( {x - 6} \right) + 2\left( {x - 6} \right)} \right) > 0\]
Thus, we get
\[\left( {\left( {x - 6} \right)\left( {x + 2} \right)} \right) > 0\]
That means \[\left( {x - 6} \right) > 0\] and \[\left( {x + 2} \right) > 0\]
So, we get
\[x > 0 + 6\] and \[x < 0 - 2\]
\[x > 6\] and \[x < - 2\]
Hence, the function \[f\] defined by \[f\left( x \right) = {x^3} - 6{x^2} - 36x + 7\] is increasing, if \[x > 6\] and \[x < - 2\]

Therefore, the correct option is (D).

Note: Many students generally make mistakes in finding the values of x. They may write it as \[x > - 2\] instead of \[x < - 2\] so that the end result will be wrong.