The interval in which the function f(x) = x3 - 6x2 + 9x + 10 is increasing in

Question

The interval in which the function f(x) = x3 - 6x2 + 9x + 10 is increasing in (A) [1, 3] (B) ( - ∞ , 1) ∪ (3, ∞) (C) ( - ∞ , - 1] ∪ [3, ∞) (D) ( - ∞ , 1] ∪ [3, ∞)

Tardigrade Admin · Accepted Answer

f(x)=x3-6x2+9x+10 f '(x)=3x2-12x+9 f '(x)=0 3x2-12x+9=0 (dividing by 3) x2-4x+3=0 x2-3x-x+3=0 x(x-3)-1(x-3)=0 x=3, 1 (-∞, 1 ] ∪[3,∞)

Q. The interval in which the function $f (x) = x^{3} - 6 x^{2} + 9 x + 10$ is increasing in

[1, 3]

$(- \infty, 1) \cup (3, \infty)$

$(- \infty, - 1] \cup [3, \infty)$

$(- \infty, 1] \cup [3, \infty)$

$f (x) = x^{3} - 6 x^{2} + 9 x + 10$
$f^{'} (x) = 3 x^{2} - 12 x + 9$
$f^{'} (x) = 0$
$3 x^{2} - 12 x + 9 = 0$ (dividing by $3$ )
$x^{2} - 4 x + 3 = 0$
$x^{2} - 3 x - x + 3 = 0$
$x (x - 3) - 1 (x - 3) = 0$
$x = 3, 1$
$(- \infty, 1] \cup [3, \infty)$

Q. The interval in which the function f(x)=x3−6x2+9x+10 is increasing in

[1, 3]

(−∞,1)∪(3,∞)

(−∞,−1]∪[3,∞)

(−∞,1]∪[3,∞)

f(x)=x3−6x2+9x+10 f′(x)=3x2−12x+9 f′(x)=0 3x2−12x+9=0(dividing by 3) x2−4x+3=0 x2−3x−x+3=0 x(x−3)−1(x−3)=0 x=3,1 (−∞,1]∪[3,∞)

Q. The interval in which the function $f (x) = x^{3} - 6 x^{2} + 9 x + 10$ is increasing in

$(- \infty, 1) \cup (3, \infty)$

$(- \infty, - 1] \cup [3, \infty)$

$(- \infty, 1] \cup [3, \infty)$

$f (x) = x^{3} - 6 x^{2} + 9 x + 10$
$f^{'} (x) = 3 x^{2} - 12 x + 9$
$f^{'} (x) = 0$
$3 x^{2} - 12 x + 9 = 0$ (dividing by $3$ )
$x^{2} - 4 x + 3 = 0$
$x^{2} - 3 x - x + 3 = 0$
$x (x - 3) - 1 (x - 3) = 0$
$x = 3, 1$
$(- \infty, 1] \cup [3, \infty)$