Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
(i) f (x) = x³, x ∈ [- 2, 2]
(ii) f (x) = sin x + cos x, x ∈ [0, π]
(iii) f (x) = 4x - 1/2 x², x ∈
(iv) f (x) = (x - 1)² + 3, x ∈ [- 3, 1]

Solution:

Maxima and minima are the maximum or the minimum value of a function within the given set of ranges.

(i) The given function is f (x) = x³

Therefore,

f' (x) = 3x²

Now,

f' (x) = 0

⇒ 3x² = 0

Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [- 2, 2].

Therefore,

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

Therefore,

f' (x) = cos x - sin x

Now,

f' (x) = 0

⇒ cos x - sin x = 0

⇒ sin x = cos x

⇒ tan x = 1

⇒ x = π/4

Then, we evaluate the value of f at critical point x = π/4 and at the end points of the interval [0, π].

Therefore,

f(π/4) = sin π / 4 + cos π / 4

= 1/√2 + 1/√2

= 2/√2

= √2

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 = π / 4.

Also, the absolute minimum value of f on [0, π] is - 1 occurring at x = π.

(iii) The given function is f (x) = 4x - 1/2 x²

Therefore,

f' (x) = 4 - 1/2 (2x)

= 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 end points of the interval [- 2, 9/2].

Therefore,

f (4) = 16 - 1/2 (16)

= 16 - 8

= 8

f (- 2) = - 8 - 1/2 (4)

= - 8 - 2

= - 10

f (9/2) = 18 - 1/2 (9/2)²

= 18 - 81/8

= 18 - 10.125

= 7.875

Hence, we can conclude that the absolute maximum value of f on [- 2, 9/2] is 8 occurring at

x = 4

Also, the absolute minimum value of f on [- 2, 9/2] is - 10 occurring at x = - 2.

(iv) The given function is f (x) = (x - 1)² + 3

Therefore,

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 end points of the interval [- 3, 1].

f (1) = (1 - 1)³ + 3

= 3

f (- 3) = (- 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.

Also, the absolute minimum value of f on [- 3, 1] is 3 occurring at x = 1

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 5

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) f (x) = x³, x ∈ [- 2, 2] (ii) f (x) = sin x + cos x, x ∈ [0, π] (iii) f (x) = 4x - 1/2 x², x ∈
(iv) f (x) = (x - 1)² + 3, x ∈ [- 3, 1]

Summary:

i) Absolute minimum value of f on [- 2, 2] is - 8 occurring at x = - 2. ii) Absolute minimum value of f on [0, π] is - 1 occurring at x = π iii) Absolute minimum value of f on [- 2, 9/2] is - 10 occurring at x = - 2