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प्रश्न
Find the local extrema for the following functions using second derivative test:
f(x) = – 3x5 + 5x3
उत्तर
f(x) = – 3x5 + 5x3
f'(x) = 0, f”(x) = – ve at x = a
⇒ x = a is a maximum point
f'(x) = 0, f”(x) = + ve at x = 6
⇒ x = b is a minimum point
f(x) = – 3x5 + 5x3
f’(x) = – 15x4 + 15x2
f”(x) = – 60x3 + 30x
f'(x) = 0
⇒ – 15x2 (x2 – 1) = 0
⇒ x = 0, +1, – 1
At x = 0, f”(x) = 0
At x = 1, f”(x) = – 60 + 30 = – ve
At x = – 1, f”(x) = 60 – 30 = + ve
So at x = 1, f'(x) = 0 and f”(x) = – ve
⇒ x = 1 is a local maximum point.
And f(1) = 2
So the local maximum is (1, 2)
At x = – 1, f'(x) = 0 and f”(x) = + ve
⇒ x = – 1 is a local maximum point and f(– 1) = – 2.
So the local minimum point is (– 1, – 2)
∴ a local minimum is – 2 and the local maximum is 2.
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