Prove that the function f(x) = x^{3} – 6x^{2} + 12x – 18 is increasing on R.

Given:- Function f(x) = x^{3} – 6x^{2} + 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) = x^{3} – 6x^{2} + 12x – 18

⇒

⇒ f’(x) = 3x^{2} – 12x + 12

⇒ f’(x) = 3(x^{2} – 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

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