Final answer:
The local linear approximation of ∛(64+x) at x₀=0 is used to approximate ∛62.5 with an approximate value of 3.96875, which is very close to the exact value when calculated.
Step-by-step explanation:
To approximate ∛62.5 using the local linear approximation of f(x) = ∛(64+x) at x₀ = 0, we first need to calculate the derivative of f(x) at x₀. To find the derivative, we can use the power rule on the function f(x) = (64+x)¹⁄₃. The derivative is f'(x) = ¹⁄₃(64+x)⁻²⁄₃. When x₀ = 0, f'(0) is ¹⁄₃(64)⁻²⁄₃, which simplifies to ¹⁄₃(4)⁻² = 1/48. The local linear approximation at x₀ = 0 is then f(x) ≈ f(0) + f'(0)(x - 0), which becomes ∛64 + (1/48)x.
To approximate ∛62.5, we substitute x = -1.5 (since 64 - 1.5 = 62.5) into the local linear approximation. The approximation yields ∛64 - (1/48)(1.5) = 4 - (1/32), which simplifies to 4 - 0.03125 = 3.96875.
Comparing this with the exact value of ∛62.5, calculated using a calculator or other numerical methods, yields the value 3.968626967, which is very close to the approximation.