Given:- Function f(x) = f(x) = x² – 6x + 3

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) = f(x) = x² – 6x + 3

⇒

⇒ f’(x) = 2x – 6

⇒ f’(x) = 2(x – 3)

Here A function is said to be increasing on [a,b] if f(x) > 0

as given

x ϵ [4, 6]

⇒ 4 ≤ x ≤ 6

⇒ 1 ≤ (x–3) ≤ 3

⇒ (x – 3) > 0

⇒ 2(x – 3) > 0

⇒ f’(x) > 0

Hence, condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ [4, 6]