University:
East Georgia State College
Course:
MATH 1540 | Calculus I
Academic year:

2022
Views:

373

Pages:

1
Author:

StellarStrategist

AP Calculus BC - Graded HW Name: Key Review for Chapter 4 and 9.1 & 9.2 a) Give the first 4 terms of a power series that represents 1/(1+x³). 1/(1+x³) = 1 - x³ + x⁶ - x⁹ b) Find the interval of convergence for your power series. Converges when |x³| < 1 |x| < 1 -1 < x < 1 Let f be a function that has derivatives of all orders for all real numbers. We know that f(3) = 1, f'(3) = 4, f''(3) = 6, and f'''(3) = 12. a) Write the third order Taylor polynomial for f at x = 3 and use it to approximate f(3.2). P₃(x) = 1 + 4(x-3) + 3(x-3)² + 2(x-3)³ P₃(3.2) = 1 + 4(0.2) + 3(0.2)² + 2(0.2)³ ≈ 1.936 b) Write the second order Taylor polynomial for f' at x = 3 and use it to approximate f'(2.7). Take derivative of P₃(x): 4 + 6(x-3) + 6(x-3)² f'(2.7) ≈ 4 + 6(-0.3) + 6(-0.3)² ≈ 2.74 Does the linearization of f underestimate or overestimate the values of f(x) near x = 3? Justify your answer. If f''(x) is concave up near 3, linearization will underestimate. If f''(x) is concave down near 3, linearization will overestimate. Since f''(3) = 6, f(x) is concave up, so linearization will underestimate. Use differentials to approximate √8.25. Use y = x^(1/2) at x = 9 to find dy. dy = (1/2)x^(-1/2) dx = (1/6)(0.25) = 1/24 ≈ 0.042 So, √8.25 ≈ √9 + 0.042 ≈ 3.042 Find the extreme values of f(x) and where they occur. f(x) = 1/√(1-x²) Domain: (-1, 1) f'(x) = x/(1-x²)^(3/2) Critical points: f'(x) = 0 when x = 0 f'(x) never undefined on D: (-1, 1) No endpoints to test. f(0) = 1/√(1-0²) = 1 Absolute min of 1 at x = 0.