Final answer:
To expand the function f(x) = e^x using a third-order Taylor series, we first find the expansion formula and then evaluate the function at x = 0.5. The value of the function at x = 0.5, expanded using a third-order Taylor series, is approximately e - e/2 + (e/8) - (e/48).
Step-by-step explanation:
To find the value of the function f(x) = e^x using a Taylor series expansion, we can write the third-order Taylor series expansion as:
f(x) = f(x₀) + f'(x₀)(x - x₀) + (f''(x₀)/2!)(x - x₀)² + (f'''(x₀)/3!)(x - x₀)³
Since the expansion point is x₀ = 1, we have f(1) = e^1 = e. Differentiating the function, we get f'(x) = e^x, f''(x) = e^x, and f'''(x) = e^x.
Plugging these values into the expansion formula, we get:
f(x) = e + e(x - 1) + (e/2)(x - 1)² + (e/6)(x - 1)³
To evaluate the function at x = 0.5, we substitute x = 0.5 into the expansion:
f(0.5) = e + e(0.5 - 1) + (e/2)(0.5 - 1)² + (e/6)(0.5 - 1)³
Simplifying this expression gives:
f(0.5) ≈ e - e/2 + (e/8) - (e/48)
Therefore, the value of the function f(x) = e^x at x = 0.5, expanded using a third-order Taylor series, is approximately e - e/2 + (e/8) - (e/48).