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**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.

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