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Prove that the function f(x) = x3 – 6x2 + 12x – 18 is increasing on R.
Given:- Function f(x) = x3 – 6x2 + 12x – 18
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) = x3 – 6x2 + 12x – 18
⇒
⇒ f’(x) = 3x2 – 12x + 12
⇒ f’(x) = 3(x2 – 4x + 4)
⇒ f’(x) = 3(x – 2)2
as given
x ϵ R
⇒ (x – 2)2> 0
⇒ 3(x – 2)2 > 0
⇒ f’(x) > 0
Hence, condition for f(x) to be increasing
Thus f(x) is increasing on interval x ∈ R