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

Last updated date: 06th Sep 2024
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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.