Given f (x) = x4 – 62x2 + 120x + 9
∴ f'(x) = 4x3 – 124x + 120 = 4(x3 – 31x + 30)
f”(x) = 12x2 – 124 = 4(3x2 – 31)
For maxima and minima, f'(x) = 0
4(x3 – 31x + 30) = 0
So roots will be x = 5, 1, – 6
Now, f”(5) = 176 > 0
x = 5 is point of local minima
f”(1) = – 112 < 0
x = 1 is point of local maxima
f”(– 6) = 308 > 0
x = – 6 is point of local minima
Local max value = f (1) = 68Answered by Aaryan | 1 year ago
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