Given that y=x^ (1/3), use calculus to determine an approximate value for ∛8030

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