Home / AP Calculus BC: 10.11 Finding Taylor Polynomial Approximations of Functions – Exam Style questions with Answer- FRQ

AP Calculus BC: 10.11 Finding Taylor Polynomial Approximations of Functions – Exam Style questions with Answer- FRQ

Question

x11.11.21.31.4
f'(x)810121314.5

The function f is twice differentiable for x > 0 with f(1) = 15 and f”(1) = 20. Values of f’, the derivative of f, are given for selected values of x in the table above.
(a) Write an equation for the line tangent to the graph of f at x =1. Use this line to approximate f(1.4).
(b) Use a midpoint Riemann sum with two subintervals of equal length and values from the table to approximate \(\) Use the approximation for \(\) to estimate the value of f(1.4). Show
the computations that lead to your answer.
(c) Use Euler’s method, starting at x = 1 with two steps of equal size, to approximate f(1.4). Show the computations that lead to your answer.
(d) Write the second-degree Taylor polynomial for f about x = 1. Use the Taylor polynomial to approximate f(1.4).

Answer/Explanation

Ans:

(a) f(1) = 15, f'(1) = 8

An equation for the tangent line is
y  = 15 + 8 (x – 1) .

f(1.4) ≈ 15 + 8 (1.4 – 1) = 18.2

(b) \(\int_{1}^{1.4}f'(x)dx\approx (0.2)(10)+(0.2)(13)=4.6\)

\(f(1.4)=f(1)+\int_{1}^{1.4}f'(x)dx\)

f(1.4) ≈ 15 + 4.6 = 19.6

(c) f(1.2) ≈ f(1) + (0.2) (8) = 16.6

f(1.4) ≈ 16.6 + (0.2) (12) = 19.0

(d) \(T_{2}(x)=15+8(x-1)+\frac{20}{2!}(x-1)^{2}\)

= 15 + 8 (x-1) + 10(x – 1)2

f(1.4) ≈ 15 + 8(1.4-1) + 10(1.4 – 1)2 = 19.8

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