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प्रश्न
Find the maximum value of 2x3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1].
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उत्तर
f(x) = 2x3 - 24x + 107, interval [1, 3]
f'(x) = 6x2 - 24
For the highest and lowest values, f‘(x) = 0
⇒ 6x2 - 24 = 0
⇒ 6x2 = 24
⇒ x2 = 4
⇒ x = ± 2
Putting the values of x in f(x) = 2x3 - 24x + 107 for the interval [1, 3],
At, x = 1 f(1) = 2(1)3 - 24 (1) + 107 = 2 - 24 + 107 = 85
At, x = 3 f (3) = 2(3)3 - 24 (3) + 107 = 54 - 72 + 107 = 89
At, x = 2 f(2) = 2(2)3 - 24(2) + 107 = 16 - 48 + 107 = 75
Thus, the maximum value of f(x) = 89,
At x = 3, for the interval [-3,-1] we find the value of f(x) at x = - 3, - 2, - 1.
At, x = – 3 f(-3) = 2(-3)3 - 24 (-3) + 107 = - 54 + 72 + 107 = - 54 + 179 = 125
At, x = – 1 f(-1) = 2 (-1)3 - 24 (-1) + 107 = -2 +24 + 107 = 129
At, x = - 2 f(-2) = 2(-2)3 - 24 (-2) + 107 = -16 + 48 +107 = 139
Thus, the maximum value of f(x) = 139 at x = -2.
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