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