Given f (x) = x^{4} – 62x^{2 +} 120x + 9

∴ f'(x) = 4x^{3} – 124x + 120 = 4(x^{3} – 31x + 30)

f”(x) = 12x^{2} – 124 = 4(3x^{2} – 31)

For maxima and minima, f'(x) = 0

4(x^{3} – 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) = 68

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