Given that y=x^(1/3) , use calculus to determine an approximate value for ∛8030
1 answer:
F(x) = x^(1/3) f(8000) = 8000^(1/3) = 20 f'(x) = 1/3 * x^(-2/3) f'(x) = 1/(3x^(2/3)) f'(8000) = 1/(3*8000^(2/3)) = 1/(3*400) = 1/1200 Linear approximation for f(x) about x = 8000 L(x) = f(8000) + f'(8000) (x−8000) L(x) = 20 + 1/1200 (x−8000) 8030^(1/3) ≈ L(8030) = 20 + 1/1200 (8030−8000) = 20 + 30/1200 = 20 + 1/40 = 20.025
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