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The function \(f(x) = \dfrac{x^3 - 5x^2 - 8x}{3}\) is

Question and options

The function f(x)=x3−5x2−8x3f(x) = \dfrac{x^3 - 5x^2 - 8x}{3} is
  1. positive and monotonically increasing for x ∈(−∞,5−572)\in (-\infty, \frac{5-\sqrt{57}}{2}) and x ∈(5+572,+∞)\in (\frac{5+\sqrt{57}}{2}, +\infty)
  2. negative and monotonically decreasing for x ∈(−∞,5−572)\in (-\infty, \frac{5-\sqrt{57}}{2}) and x ∈(5+572,+∞)\in (\frac{5+\sqrt{57}}{2},+\infty)
  3. negative and monotonically increasing for x ∈(−∞,5−572)\in (-\infty, \frac{5-\sqrt{57}}{2}) and positive and monotonically increasing for x ∈(5+572,+∞)\in (\frac{5+\sqrt{57}}{2},+\infty)
  4. positive and monotonically increasing for x ∈(−∞,5−572)\in (-\infty, \frac{5-\sqrt{57}}{2}) and negative and monotonically decreasing for x ∈(5+572,+∞)\in (\frac{5+\sqrt{57}}{2},+\infty)

Answer and explanation

Correct answer: C - negative and monotonically increasing for x \(\in (-\infty, \frac{5-\sqrt{57}}{2})\) and positive and monotonically increasing for x \(\in (\frac{5+\sqrt{57}}{2},+\infty)\)
f = x(x^2-5x-8)/3 has roots 0 and (5+-sqrt57)/2 (about -1.27, 6.27); f' = (3x+2)(x-4)/3 > 0 for x < -2/3 and x > 4. So f is negative and increasing for x < (5-sqrt57)/2, positive and increasing for x > (5+sqrt57)/2.

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