Exercises for Introduction to Interpolation 1. Consider P(x)=-0.02x^3+0.1x^2-0.2x+1.66 which passes through the four points (1,1.54),(2,1.5),(3,1.42) and (5,0.66) (a) Find P(4) (b) Find P'(4) (c) Find the definite integral of P(x) taken over [1,4] (d) Find the extrapolated value P(5.5) (e) Show how to find the coefficients of P(x)

(a) To find $P(4)$, we substitute $x=4$ into the polynomial $P(x)$:<br /><br />$P(4) = -0.02(4)^3 + 0.1(4)^2 - 0.2(4) + 1.66$<br /><br />$P(4) = -0.02(64) + 0.1(16) - 0.2(4) + 1.66$<br /><br />$P(4) = -1.28 + 1.6 - 0.8 + 1.66$<br /><br />$P(4) = 1.18$<br /><br />(b) To find $P'(4)$, we first need to find the derivative of $P(x)$:<br /><br />$P'(x) = \frac{d}{dx}(-0.02x^3 + 0.1x^2 - 0.2x + 1.66)$<br /><br />$P'(x) = -0.06x^2 + 0.2x - 0.2$<br /><br />Now, we substitute $x=4$ into $P'(x)$:<br /><br />$P'(4) = -0.06(4)^2 + 0.2(4) - 0.2$<br /><br />$P'(4) = -0.06(16) + 0.8 - 0.2$<br /><br />$P'(4) = -0.96 + 0.8 - 0.2$<br /><br />$P'(4) = -0.36$<br /><br />(c) To find the definite integral of $P(x)$ taken over $[1,4]$, we integrate $P(x)$ with respect to $x$ from 1 to 4:<br /><br />$\int_{1}^{4} P(x) \, dx = \int_{1}^{4} (-0.02x^3 + 0.1x^2 - 0.2x + 1.66) \, dx$<br /><br />To evaluate this integral, we can use the Fundamental Theorem of Calculus. First, we find the antiderivative of $P(x)$:<br /><br />$\int P(x) \, dx = -0.005x^4 + \frac{1}{30}x^3 - 0.1x^2 + 1.66x + C$<br /><br />Now, we evaluate this antiderivative at the upper and lower limits of integration:<br /><br />$\int_{1}^{4} P(x) \, dx = \left[-0.005(4)^4 + \frac{1}{30}(4)^3 - 0.1(4)^2 + 1.66(4)\right] - \left[-0.005(1)^4 + \frac{1}{30}(1)^3 - 0.1(1)^2 + 1.66(1)\right]$<br /><br />$\int_{1}^{4} P(x) \, dx = \left[-0.32 + \frac{64}{30} - 1.6 + 6.64\right] - \left[-0.005 + \frac{1}{30} - 0.1 + 1.66\right]$<br /><br />$\int_{1}^{4} P(x) \, dx = \left[-0.32 + 2.1333 - 1.6 + 6.64\right] - \left[-0.005 + 0.0333 - 0.1 + 1.66\right]$<br /><br />$\int_{1}^{4} P(x) \, dx = 6.8533 - 1.7053$<br /><br />$\int_{1}^{4} P(x) \, dx = 5.148$<br /><br />(d) To find the extrapolated value $P(5.5)$, we substitute $x=5.5$ into the polynomial $P(x)$:<br /><br />$P(5.5) = -0.02(5.5)^3 + 0.1(5.5)^2 - 0.2(5.5) + 1.66$<br /><br />$P(5.5) = -0.02(166.375) + 0.1(30.25) - 0.2(5.5) + 1.66$<br /><br />$P(5.5) = -3.3275 + 3.025 - 1.1 + 1.66$<br /><br />$P(5.5) = -0.7325$<br /><br />(e) To find the coefficients of $P(x)$, we can use the given points to set up a system of equations. Since $P(x)$ passes through the points $(1,1.54)$, $(2,1.5)$, $(3,1.42)$, and $(5,0.66)$, we can