State when a function f(x) is said to be increasing on an interval [a, b]. Test whether the function f(x) = x2 – 6x + 3 is increasing on the interval [4, 6].

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]


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