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Maxima and Minima — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMaxima and Minima4 Marks
Question
Find the absolute maximum and the absolute minimum value of the following functions in the given intervals:
f(x) = 3x4 - 8x3 + 12x2 - 48x + 25 in [0,3]
✓

Answer

Given, f(x) = 3x4 - 8x3 + 12x2 - 48 + 25
⇒ f'(x) = 12x3 - 24x2 + 24x - 28
For a local maximum or a local minimum, We must have f'(x) = 0
⇒ 12x3 - 24x2 + 24x - 48 = 0
⇒ x3 - 2x2 + 2x - 4 = 0
⇒ x2(x - 2) + 2(x - 2) = 0
⇒ (x - 2)(x2 + 2) = 0
⇒ x - 2 = 0 or (x2 + 2) = 0
⇒ x = 2
No, real root exists for (x2 + 2) = 0
Thus, the critical points of f are 0, 2 and 3.
Now, f(0) = 3(0)4 - 8(0)3 + 12(0)2 - 48(0) + 25 = 25
f(2) = 3(2)4 - 8(2)3 + 12(2)2 - 48(2) + 25 = -39
f(3) = 3(3)4 - 8(3)3 + 12(3)2 - 48(3) + 25 = 1
Hence, the absolute maximum value when x = 0 is 25 and the absolute minimum value when x = 2 is −39.

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