Q. 1 D

# Find the interval

Given:- Function f(x) = 2x3 – 12x2 + 18x + 15

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 – 12x2 + 18x + 15 f’(x) = 6x2 – 24x + 18

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

f’(x) = 0

6x2 – 24x + 18 = 0

6(x2 – 4x + 3) = 0

6(x2 – 3x – x + 3) = 0

6(x – 3)(x – 1) = 0

(x – 3)(x – 1) = 0

x = 3 , 1

clearly, f’(x) > 0 if x < 1 and x > 3

and f’(x) < 0 if 1< x < 3

Thus, f(x) increases on (–,1) (3, ∞)

and f(x) is decreasing on interval x (1,3)

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