Question

The value of $x$ for which $f(x) = x^3 - 6x^2 - 36x + 7 $ is increasing, belong to

• A. $(-1, 0) \cup (1, 5)4$
• B. $(-2, 0) \cup (1, 6)$
• C. $(\infty, 2) \cup (6, \infty)$
• D. $(-2, 6)$

Answer 1

Answer: (C)

$f\left(x\right)=x^{3 } -6x^{2} -36x+7$
$ f'\left(x\right)= 3x^{2}-12x -36 $
Now, $f'\left(x\right) = 0$
$ \Rightarrow 3x^{2} - 12x -36 = 0 $
$\Rightarrow x=-2,6 $
Hence,f(x) is increasing in $(\infty 2) \cup (6, \infty)$