Find the integral of the function f(x)= x^3 - 12x^2 + 36 x + 17.(i) Minima (ii) Maxima.
Answers
Answer:
Given:- Function f(x) = x³ – 12x² + 36x + 17
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)
x€(a, b)
(ii) Find f’(x)
x€(a, b)
(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) = x³ – 12x² + 36x + 17
⇒
f(x)=d/dx(x³-12x²+36x+17)
⇒ f’(x) = 3x² – 24x + 36
For f(x) lets find critical point, we must have
⇒ f’(x) = 0
⇒ 3x² – 24x + 36 = 0
⇒ 3(x² – 8x + 12) = 0
⇒ 3(x² – 6x – 2x + 12) = 0
⇒ x² – 6x – 2x + 12 = 0
⇒ (x – 6)(x – 2) = 0
⇒ x = 2, 6
clearly, f’(x) > 0 if x < 2 and x > 6
and f’(x) < 0 if 2 < x < 6
Thus, f(x) increases on (–∞, 2) ∪ (6, ∞)
and f(x) is decreasing on interval x ∈ (2, 6)
Answer:
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