Question

$2x^{3} + 18x^{2} - 96x + 45 = 0$ is an increasing function when

• A. $x \leq - 8,x \geq 2$
• B. $x < - 2,x \geq 8$
• C. $x \leq - 2,x \geq 8$
• D. $0 < x \leq - 2$

Answer 1

Answer: (A)

$x \leq - 8,x \geq 2$

Answer 2

$f^{'}(x) = 6x^{2} + 36x - 96 > 0$, for increasing

⇒ $f^{'}(x) = 6(x + 8)(x - 2) \geq 0$ ⇒ $x \geq 2,x \leq - 8$.