The function f (x)=2x3-15x2 +36x +6 is strictly decreasing in the interval

Question

The function f (x)=2x3-15x2 +36x +6 is strictly decreasing in the interval (A) (2, 3) (B) (-∞, 2) (C) (3, 4) (D) (-∞, 3)∪(4, ∞)

Tardigrade Admin · Accepted Answer

Given, f(x)=2 x3-15 x2+36 x+6 On differentiating both sides w.r.t. x, we get f'(x)=6 x2-30 x+36 For strictly decreasing, f'(x)<0 ⇒ 6(x2-5 x+6)<0 ⇒ (x-3)(x-2)<0 ⇒ 2 < x < 3 ∴ x ∈(2,3)

Q. The function $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 6$ is strictly decreasing in the interval

(2, 3)

$(- \infty, 2)$

(3, 4)

$(- \infty, 3) \cup (4, \infty)$

$(- \infty, 2) \cup (3, \infty)$

Given, $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 6$
On differentiating both sides w.r.t. $x$ , we get
$f^{^{'}} (x) = 6 x^{2} - 30 x + 36$
For strictly decreasing,
$f^{^{'}} (x) < 0$
$\Rightarrow 6 (x^{2} - 5 x + 6) < 0$
$\Rightarrow (x - 3) (x - 2) < 0$
$\Rightarrow 2 < x < 3$
$∴ x \in (2, 3)$

Q. The function f(x)=2x3−15x2+36x+6 is strictly decreasing in the interval

(2, 3)

(−∞,2)

(3, 4)

(−∞,3)∪(4,∞)

(−∞,2)∪(3,∞)

Given, f(x)=2x3−15x2+36x+6 On differentiating both sides w.r.t. x, we get f′(x)=6x2−30x+36 For strictly decreasing, f′(x)<0 ⇒6(x2−5x+6)<0 ⇒(x−3)(x−2)<0 ⇒2<x<3 ∴x∈(2,3)

Q. The function $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 6$ is strictly decreasing in the interval

$(- \infty, 2)$

$(- \infty, 3) \cup (4, \infty)$

$(- \infty, 2) \cup (3, \infty)$

Given, $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 6$
On differentiating both sides w.r.t. $x$ , we get
$f^{^{'}} (x) = 6 x^{2} - 30 x + 36$
For strictly decreasing,
$f^{^{'}} (x) < 0$
$\Rightarrow 6 (x^{2} - 5 x + 6) < 0$
$\Rightarrow (x - 3) (x - 2) < 0$
$\Rightarrow 2 < x < 3$
$∴ x \in (2, 3)$