\begin{tabular}{l}\hline Exercise 4.14 \\ Consider data points \( (0,2),(1,5) \) and \( (2,25) \). Find an interpolating polynomial \( p(x) \) \\ of at most degree two and estimate the values of \( p(-1) \) and \( \rho(3) \). \\ \end{tabular}
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To find an interpolating polynomial \( p(x) \) of at most degree two for the points \( (0,2), (1,5), (2,25) \), we can use the Lagrange interpolation formula or the form of a quadratic polynomial \( p(x) = ax^2 + bx + c \). Let's use the points to set up the system of equations: 1. For \( (0,2) \): \( p(0) = c = 2 \) 2. For \( (1,5) \): \( p(1) = a(1)^2 + b(1) + c = a + b + 2 = 5 \) 3. For \( (2,25) \): \( p(2) = a(2)^2 + b(2) + c = 4a + 2b + 2 = 25 \) Now, we solve the equations: From equation 2: \[ a + b + 2 = 5 \] \[ a + b = 3 \] (equation 4) From equation 3: \[ 4a + 2b + 2 = 25 \] \[ 4a + 2b = 23 \] \[ 2a + b = 11.5 \] (equation 5) Now we have two equations (4) and (5): 1. \( a + b = 3 \) 2. \( 2a + b = 11.5 \) Subtract equation (4) from (5): \[ (2a + b) - (a + b) = 11.5 - 3 \] \[ a = 8.5 \] Substituting \( a = 8.5 \) back into equation (4): \[ 8.5 + b = 3 \] \[ b = 3 - 8.5 = -5.5 \] So, we have: - \( a = 8.5 \) - \( b = -5.5 \) - \( c = 2 \) Thus, the polynomial is: \[ p(x) = 8.5x^2 - 5.5x + 2 \] Now we can estimate the values of \( p(-1) \) and \( p(3) \): For \( p(-1) \): \[ p(-1) = 8.5(-1)^2 - 5.5(-1) + 2 = 8.5 + 5.5 + 2 = 16 \] For \( p(3) \): \[ p(3) = 8.5(3)^2 - 5.5(3) + 2 = 8.5(9) - 16.5 + 2 = 76.5 - 16.5 + 2 = 62 \] Thus, the interpolating polynomial is: \[ p(x) = 8.5x^2 - 5.5x + 2 \] And the estimates are: \[ p(-1) = 16 \] \[ p(3) = 62 \]
