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Find the Maximum Value of 2x3 − 24x + 107 in the Interval [1, 3]. Find the Maximum Value of the Same Function in [−3, −1]. - CBSE (Science) Class 12 - Mathematics

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Question

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

Let f(x) = 2x3 − 24x + 107.

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].

f(2) = 2(8) − 24(2) + 107 = 16 − 48 + 107 = 75

f(1) = 2(1) − 24(1) + 107 = 2 − 24 + 107 = 85

f(3) = 2(27) − 24(3) + 107 = 54 − 72 + 107 = 89

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

Next, we consider the interval [−3, −1].

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

f(−3) = 2 (−27) − 24(−3) + 107 = −54 + 72 + 107 = 125

f(−1) = 2(−1) − 24 (−1) + 107 = −2 + 24 + 107 = 129

f(−2) = 2(−8) − 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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APPEARS IN

 NCERT Solution for Mathematics Textbook for Class 12 (2018 to Current)
Chapter 6: Application of Derivatives
Q: 10 | Page no. 232
Solution Find the Maximum Value of 2x3 − 24x + 107 in the Interval [1, 3]. Find the Maximum Value of the Same Function in [−3, −1]. Concept: Maxima and Minima.
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