6) Find the intervals in which the following functions are strictly increasing or
decreasing:

c) - 2 x^3 - 9x ^2 - 12 x + 1

Answers (1)

Given function is,
f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{}
f^{'}(x) = - 6 x^2 - 18x - 12
Now,
f^{'}(x) = 0\\ - 6 x^2 - 18x - 12 = 0\\ -6(x^{2}+3x+2) = 0 \\ x^{2}+3x+2 = 0 \\x^{2} + x + 2x + 2 = 0\\ x(x+1) + 2(x+1) = 0\\ (x+2)(x+1) = 0\\ x = -2 \ and \ x = -1

So, the range is (-\infty , -2) \ , (-2,-1) \ and \ (-1,\infty)
In interval  (-\infty , -2) \cup \ (-1,\infty)  , f^{'}(x) = - 6 x^2 - 18x - 12 is -ve
Hence, f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{} is strictly decreasing in interval  (-\infty , -2) \cup \ (-1,\infty)
In interval (-2,-1)  , f^{'}(x) = - 6 x^2 - 18x - 12 is +ve
Hence, f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{} is strictly increasing in the interval (-2,-1)

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