Does Taylor's Theorem with Remainder guarantee that the second Taylor polynomial of f(x)=8ex at x=2 has an error less than 0.1 in the estimate of 8e2.3 ?

To answer this question, we need to first understand what Taylor's Theorem with Remainder states. The theorem provides an approximation of a function near a point based on the function's derivatives at that point. The error in the approximation is given by the remainder term.

The second Taylor polynomial for the function f(x) = 8e^x at x = 2 is given by:

P_2(x) = f(2) + f'(2)(x - 2) + f''(2)(x - 2)^2 / 2!

We know that f(x) = 8e^x, so f'(x) = 8e^x and f''(x) = 8e^x. Therefore,

P_2(x) = 8e^2 + 8e^2(x - 2) + 8e^2(x - 2)^2 / 2

We want to estimate the value of the function at x = 2.3, so we substitute x = 2.3 into the polynomial to get:

P_2(2.3) = 8e^2 + 8e^2(2.3 - 2) + 8e^2(2.3 - 2)^2 / 2

The error in the approximation is given by the absolute difference between the actual value of the function at x = 2.3 and the estimated value using the Taylor polynomial:

|f(2.3) - P_2(2.3)|

The remainder term in Taylor's Theorem gives an upper bound for this error. For the second Taylor polynomial, the remainder term is given by:

R_2(x) = f'''(c)(x - 2)^3 / 3!

for some c between 2 and x. Since f'''(x) = 8e^x, which is always positive for x > 0, the maximum value of the remainder term occurs at x = 2.3. Therefore, the error in the approximation is less than:

|8e^2.3(2.3 - 2)^3 / 3!|

We want this error to be less than 0.1. Therefore, we need to check whether:

|8e^2.3(2.3 - 2)^3 / 3!| < 0.1

If this inequality holds, then Taylor's Theorem with Remainder guarantees that the second Taylor polynomial of f(x) = 8e^x at x = 2 has an error less than 0.1 in the estimate of 8e^2.3. If the inequality does not hold, then the theorem does not provide such a guarantee.