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Question

Find the values of x, for which the function f(x)=x3+12x2+36x+6 is increasing.

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Solution

Let
f(x)=x3+12x2+36x+6
Differentiate it w.r.t x
f(x)=3x2+24x+36
The function increases when f(x)>0
So,
3x2+24x+36>0
x2+8x+12>0
x2+6x+2x+12>0
x(x+6)+2(x+6)>0
(x+2)(x+6)>0
x<2
x<6
So, for x<2 and x<6 the function is increasing\

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