The value of (x) for which (f(x) = x^3 - 6x^2 - 36x + 7 ) is... | Mathematics | SolveeIt

Question

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

$(-1, 0) \cup (1, 5)4$

$(-2, 0) \cup (1, 6)$

$(\infty, 2) \cup (6, \infty)$

$(-2, 6)$

Answer

$(\infty, 2) \cup (6, \infty)$

Explanation

Solution

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

Hence,f(x) is increasing in $(\infty 2) \cup (6, \infty)$

Mathematics Question on Application of derivatives

Solution