Find the maximum value of 2x³ - 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [- 3, - 1]

Solution:

Maxima and minima are known as the extrema of a function.

Let f (x) = 2x³ - 24x + 107

Therefore,

On differentiating wrt x, we get

f' (x) = 6x² - 24

= 6(x² - 4)

Now,

f' (x) = 0

⇒ 6(x² - 4) = 0

⇒ x² = 4

⇒ x = ± 2

We first consider the interval [1, 3].

Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the end points of the interval [1, 3].

Hence,

f (2) = 2 (2)³ - 24 (2) + 107

= 16 - 48 + 107

= 75

f (1) = 2 (1)³ - 24 (1) + 107

= 2 - 24 + 107

= 85

f (3) = 2 (3)³ - 24 (3) + 107

= 54 - 72 + 107

= 89

Thus, the absolute maximum value of f (x) in the interval [1, 3] is 89 occurring at x = 3.

Next, we first consider the interval [- 3, - 1].

Then, we evaluate the value of f at the critical point x = - 2 ∈ [- 3, - 1] and at the end points of the interval [- 3, - 1].

Hence,

f (- 3) = 2 (- 3)³ - 24 (- 3) + 107

= - 54 + 72 + 107

= 125

f (- 1) = 2 (- 1)³ - 24 (- 1) + 107

= - 2 + 24 + 107

= 129

f (- 2) = 2 (- 2)³ - 24 (- 2) + 107

= - 16 + 48 + 107

= 139

Hence, the absolute maximum value of f (x) in the interval [- 3, - 1] is 139 occurring at x = - 2

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 10

Find the maximum value of 2x³ - 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [- 3, - 1].

Summary:

The absolute maximum value of f (x) in the interval [- 3, - 1] is 139 occurring at x = - 2. The absolute maximum value of f (x) in the interval [1, 3] is 89 occurring at x = 3