Question
Function f(x) = x3 - 27x + 5 is monotonically increasing when:
  1. $\text{x}<-3$
  2. $|\text{x}|>3$
  3. $\text{x}\leq-3$
  4. $|\text{x}|\geq3$

Answer

  1. $|\text{x}|\geq3$

Solution:

f(x) = 3x2 - 27x

⇒ f'(x) = x3 - 27x + 5

⇒ f'(x) = 3(x2 - 9)

Function is increasing,

$3\big(\text{x}^2-9\big)\geq0$

$\Rightarrow\text{x}^2\geq9$

$\Rightarrow|\text{x}|\geq3$

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