1 Use Taylor method to linearize f(x) = -3x3 + 30x when x = -3.

2.1 Use Taylor method to linearize f(x) = -3x^3 + 30x when x = -3. The Taylor linearization (first-order Taylor polynomial) of f(x) around x=a is given by L(x) = f(a) + f'(a)(x-a). Here, f(x) = -3x^3 + 30x and a = -3. Step 1: Calculate f(a). f(-3) = -3(-3)^3 + 30(-3) f(-3) = -3(-27) - 90 f(-3) = 81 - 90 f(-3) = -9 Step 2: Calculate the first derivative f'(x). f'(x) = (d)/(dx)(-3x^3 + 30x) f'(x) = -9x^2 + 30 Step 3: Calculate f'(a). f'(-3) = -9(-3)^2 + 30 f'(-3) = -9(9) + 30 f'(-3) = -81 + 30 f'(-3) = -51 Step 4: Substitute f(a) and f'(a) into the linearization formula. L(x) = f(a) + f'(a)(x-a) L(x) = -9 + (-51)(x - (-3)) L(x) = -9 - 51(x + 3) L(x) = -9 - 51x - 153 L(x) = -51x - 162 The linearized function is -51x - 162. 2.2 If R(x) = 22 + 25(x-2) + 11(x-2)^2 + 2(x-2)^3, prove that R(x) is the Taylor polynomial for f(x) = 2x^3 - x^2 + 5x, where a = 2. The Taylor polynomial of degree 3 for f(x) centered at a=2 is given by: P_3(x) = f(2) + f'(2)(x-2) + (f''(2))/(2!)(x-2)^2 + (f'''(2))/(3!)(x-2)^3 We need to calculate the derivatives of f(x) at x=2 and compare the coefficients. Given f(x) = 2x^3 - x^2 + 5x. Step 1: Calculate f(2). f(2) = 2(2)^3 - (2)^2 + 5(2) f(2) = 2(8) - 4 + 10 f(2) = 16 - 4 + 10 f(2) = 22 This matches the constant term in R(x). Step 2: Calculate f'(x) and f'(2). f'(x) = 6x^2 - 2x + 5 f'(2) = 6(2)^2 - 2(2) + 5 f'(2) = 6(4) - 4 + 5 f'(2) = 24 - 4 + 5 f'(2) = 25 This matches the coefficient of (x-2) in R(x). Step 3: Calculate f''(x) and f''(2). f''(x) = 12x - 2 f''(2) = 12(2) - 2 f''(2) = 24 - 2 f''(2) = 22 The coefficient of (x-2)^2 in the Taylor polynomial is (f''(2))/(2!). (f''(2))/(2!) = (22)/(2) = 11 This matches the coefficient of (x-2)^2 in R(x). Step 4: Calculate f'''(x) and f'''(2). f'''(x) = 12 f'''(2) = 12 The coefficient of (x-2)^3 in the Taylor polynomial is (f'''(2))/(3!). (f'''(2))/(3!) = (12)/(6) = 2 This matches the coefficient of (x-2)^3 in R(x). Step 5: Form the Taylor polynomial P_3(x) and compare it with R(x). Substituting the calculated values: P_3(x) = 22 + 25(x-2) + 11(x-2)^2 + 2(x-2)^3 Since P_3(x) is identical to R(x), R(x) is indeed the Taylor polynomial for f(x) centered at a=2. 2.3 What makes you think that R(x) is the Taylor polynomial of f(x) in 2.2. R(x) is the Taylor polynomial of f(x) because its coefficients match the values of f(x) and its successive derivatives evaluated at x=2, divided by the corresponding factorials. This directly follows the definition of a Taylor polynomial, where each coefficient f^(k)(a)k! was verified to be correct. Send me the next one 📸