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

Asked by Aaryan | 1 year ago |  80

##### Solution :-

Let f(x) = 2x3 – 24 x + 107

∴ f'(x) = 6 x2 – 24 = 6(x2 – 4)

Now, for local maxima and local minima we have 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].

f (2) = 2 (23)– 24 (2) + 107 = 75

f (1) = 2(1)3 – 24(1) + 107 = 85

f (3) = 2(3)3– 24 (3) + 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]

f (-3) = 2 (-3)3 – 24 (-3) + 107 = 125

f (-2) = 2 (-2)3 – 24 (-3) + 107 = 139

f (-1) = 2 (-1)3 – 24 (-2) + 107 = 129

Hence, the absolute maximum value of f is 139 and occurs when x = -2.

Answered by Sakshi | 1 year ago

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