Find the absolute maximum and the absolute minimum values of the functions in the given intervals f (x) = 3x4 – 8x3 + 12x2 – 48x + 25 on [0, 3]

Asked by Aaryan | 1 year ago |  66

##### Solution :-

Given function is f(x) = 3x4 – 8x3 + 12x2 – 48x + 25 on [0, 3]

On differentiating we get

f’ (x) = 12x3 – 24x2 + 24 x – 48

f’ (x) = 12 (x3 – 2x2 + 2x – 4)

f’ (x) = 12 (x – 2) (x2 + 2)

Now, for local minima and local maxima we have f'(x) = 0

x = 2 or x2 + 2 = 0 for which there are no real roots.

Therefore, we consider only x = 2 ∈ [0, 3].

Then, we evaluate of f at critical point x = 2 and at the interval [0, 3]

f (2) = 3 (2)4 – 8 (2)3 + 12 (2)2 – 48 (2) + 25

f (2) = 48 – 64 + 48 – 96 + 25 = – 39

f (0) = 3 (0)4 – 8 (0)3 + 12 (0)2 – 48 (0) + 25 = 25

f (3) = 3 (3)4 – 8 (3)3 + 12 (3)2 – 48 (3) + 25 = 16

Hence, we can conclude that the absolute maximum value of f on [0, 3] is 25 occurring at x = 0 and the minimum value of f on [0, 3] is – 39 occurring at x = 2

Answered by Sakshi | 1 year ago

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