The value of x for which f(x) = x3 - 6x2 - 36x + 7 is increasing, belong to

Question

The value of x for which f(x) = x3 - 6x2 - 36x + 7 is increasing, belong to (A) (-1, 0) ∪ (1, 5)4 (B) (-2, 0) ∪ (1, 6) (C) (∞, 2) ∪ (6, ∞) (D) (-2, 6)

Tardigrade Admin · Accepted Answer

f(x)=x3 -6x2 -36x+7 f'(x)= 3x2-12x -36 Now, f'(x) = 0 ⇒ 3x2 - 12x -36 = 0 ⇒ x=-2,6 Hence,f(x) is increasing in (∞ 2) ∪ (6, ∞)

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

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

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

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

$(- 2, 6)$

$f (x) = x^{3} - 6 x^{2} - 36 x + 7$
$f^{'} (x) = 3 x^{2} - 12 x - 36$
Now, $f^{'} (x) = 0$
$\Rightarrow 3 x^{2} - 12 x - 36 = 0$
$\Rightarrow x = - 2, 6$
Hence,f(x) is increasing in $(\infty2) \cup (6, \infty)$

Q. The value of x for which f(x)=x3−6x2−36x+7 is increasing, belong to

(−1,0)∪(1,5)4

(−2,0)∪(1,6)

(∞,2)∪(6,∞)

(−2,6)