Listen NCERT Audio Books to boost your productivity and retention power by 2X.
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]