NCERT Solutions
Class 12 Maths
Application of Derivatives
Ex.6.5 Q.5
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
(i) f(x) = x2, x є [-2, 2]
(ii). f(x) = sin x + cos x, x є [0, π]
(iii) f(x) = 4x – x2/2, x є [-2, 9/2]
(iv) f(x) = (x - 1)2 + 3, x є [-3, 2]
(i) The given function is f(x) = x3
So, f’(x) = 3x2
Now, f’(x) = 0
⟹ 3x2 = 0
⟹ x = 0
Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [−2, 2].
f(0) = 0
f(−2) = (−2)3 = −8
f(2) = (2)3 = 8
Hence, we can conclude that the absolute maximum value of f on [−2, 2] is 8 occurring at x = 2.
Also, the absolute minimum value of f on [−2, 2] is −8 occurring at x = −2.
(ii)The given function is f(x) = sin x + cos x
So, f’(x) = cos x – sin x
Now, f’(x) = 0
⟹ cos x – sin x = 0
⟹ sin x = cos x
⟹ tan x = 1
⟹ x =
Then, we evaluate the value of f at critical point x = and at the end points of the interval [0, π].
f( ) = sin + cos = 
f(0) = sin 0 + cos 0 = 0 + 1 = 1
f(π) = sin π + cos π = 0 – 1 = -1
Hence, we can conclude that the absolute maximum value of f on [0, π] is √2
occurring at x = and the absolute minimum value of f on [0, π] is −1 occurring at x = π.
(iii) The given function is f(x) = 
So, f’(x) 
= 4 – x
Now, f’(x) = 0
⟹ 4 – x = 0
⟹ x = 4
Then, we evaluate the value of f at critical point x = 4 and at the end points of the interval [−2, 
].
Now,
f(4) = 16 – 
= 16 – 8 = 8
f(-2) = -8 – 
= -8 – 2 = -10
f( ) = 4( ) – (
)( )2 = 18 – 
= 18 – 10.125 = 7.875
Hence, we can conclude that the absolute maximum value of f on [−2, ] is
8 occurring at x = 4 and the absolute minimum value of f on [−2, ] is -10 occurring at x = -2.
(iv) The given function is f(x) = (x - 1)2 + 3
So, f’(x) = 2(x - 1)
Now, f’(x) = 0
⟹ 2(x – 1) = 0
⟹ x = 1
Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval
[−3, 1].
Now,
f(1) = (1 - 1)2 + 3 = 0 + 3 = 3
f(1) = (-3 - 1)2 + 3 = 16 + 3 = 19
Hence, we can conclude that the absolute maximum value of f on [−3, 1] is 19 occurring at x = -3
and the absolute minimum value of f on [−3, 1] is 3 occurring at x = 1.