Let's answer **part (a)** and **part (b)** step-by-step, with explanations at the end of each step. We'll find the Taylor polynomial (step 1), then use it to approximate values (step 2). --- ### **Step 1: Find the Taylor polynomial of degree 4 for \( f(x) = x^2 - x \) about \( x_0 = 1 \).** Recall the general Taylor polynomial of degree 4 centered at \( x_0 \): \[ P_4(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \frac{f'''(x_0)}{3!}(x-x_0)^3 + \frac{f^{(4)}(x_0)}{4!}(x-x_0)^4 \] First, compute the derivatives: - \( f(x) = x^2 - x \) - \( f'(x) = 2x - 1 \) - \( f''(x) = 2 \) - \( f'''(x) = 0 \) - \( f^{(4)}(x) = 0 \) Now, evaluate at \( x_0 = 1 \): - \( f(1) = 1^2 - 1 = 0 \) - \( f'(1) = 2 \cdot 1 - 1 = 1 \) - \( f''(1) = 2 \) - \( f'''(1) = 0 \) - \( f^{(4)}(1) = 0 \) Plug into the Taylor formula: \[ P_4(x) = 0 + 1(x-1) + \frac{2}{2!}(x-1)^2 + 0 + 0 \] \[ = (x-1) + (x-1)^2 \] **Explanation (Step 1):** We calculated the first four derivatives of \( f(x) \) and evaluated them at \( x_0 = 1 \). Since the third and fourth derivatives are zero, their terms vanish. The Taylor polynomial of degree 4 is \( \boxed{P_4(x) = (x-1) + (x-1)^2} \). --- ### **Step 2: Use the result of (a) to approximate \( f(0.9) \) and \( f(1.1) \).** Substitute \( x = 0.9 \) and \( x = 1.1 \) into \( P_4(x) \): #### For \( x = 0.9 \): \[ P_4(0.9) = (0.9-1) + (0.9-1)^2 = (-0.1) + (-0.1)^2 = -0.1 + 0.01 = -0.09 \] #### For \( x = 1.1 \): \[ P_4(1.1) = (1.1-1) + (1.1-1)^2 = (0.1) + (0.1)^2 = 0.1 + 0.01 = 0.11 \] **Explanation (Step 2):** We substituted the values \( x = 0.9 \) and \( x = 1.1 \) into the Taylor polynomial to get the approximations \( f(0.9) \approx -0.09 \) and \( f(1.1) \approx 0.11 \). --- ### **Final Answers:** - **(a) The Taylor polynomial of degree 4 is \( \boxed{(x-1) + (x-1)^2} \).** - **(b) Using this polynomial:** - \( f(0.9) \approx \boxed{-0.09} \) - \( f(1.1) \approx \boxed{0.11} \)