5) Find the intervals in which the function f given by f (x) = 2x^3 - 3x ^2 - 36 x + 7 is
 a) increasing 

Answers (1)

It is given that
f (x) = 2x^3 - 3x ^2 - 36 x + 7
So,
f^{'}(x)= 6x^{2} - 6x - 36
f^{'}(x)= 0
6x^{2} - 6x -36 =0 \Rightarrow 6 (x^{2} - x-6)
x^{2} - x-6 = 0
x^{2} - 3x+2x-6 = 0
x(x-3) + 2(x-3) = 0\\
(x+2)(x-3) = 0
x = -2 ,  x = 3

So, three ranges are there   (-\infty,-2) , (-2,3) \ and \ (3,\infty)
Function    f^{'}(x)= 6x^{2} - 6x - 36  is positive in interval  (-\infty,-2) , (3,\infty)   and negative in the interval  (-2,3)
Hence, f (x) = 2x^3 - 3x ^2 - 36 x + 7  is strictly increasing in (-\infty,-2) \cup (3,\infty)
and  strictly decreasing in the interval  (-2,3) 

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