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# Find the interval

Answer :

Given:- Function f(x) = -2x3 + 3x2 + 12x + 6

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = –2x3 + 3x2 + 12x + 6 f’(x) = –6x2 + 6x + 12

For f(x) lets find critical point, we must have

f’(x) = 0

–6x2 + 6x + 12 = 0

6(–x2 + x + 2) = 0

6(–x2 + 2x – x + 2) = 0

x2 – 2x + x – 2 = 0

(x – 2)(x + 1) = 0

x = –1, 2

clearly, f’(x) > 0 if –1 < x < 2

and f’(x) < 0 if x < –1 and x > 2

Thus, f(x) increases on x (–1, 2)

and f(x) is decreasing on interval (–, –1) (2, ∞)

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