Linear approximation and (the second derivative) Draw the graph of a function ?f? such that ?f?(1) = ?f'?(1) = ?f"?(1) = 1,Draw the linear approximation to the function at the point (1, 1). Now draw the graph of another function g such that g(1) = ?g??(1) = 1 and ?g??(1) = 10, (It is not possible to represent the second derivative exactly, but your graphs should reflect the fact that ?f"?(1) is relatively small and ?g"?(1) is relatively large.) Now, suppose that linear approximations are used to approximate ?f( ? 1.1) and g(1.1). a. Which function has the more accurate linear approximation near ?x? = 1 and why? b. Explain why the error in the linear approximation to ?f? near a point ?a? is proportional to the magnitude of ?f"?(?a?).

SOLUTION STEP 1 Here it is given that f(1)(1) = (1) = 1 Thus we can understand that a=1 .The linear approximation formula is given by L(x) = f(a)+f(a)(xa) L(x) = 1+1(x1) = 1+x1 = x Thus we have L(x) = x STEP 2 The graph will be Here the black line is the actual curve and the other line is the linear approximation.since f"(1) = 1 is not very large,the change in slope will be less gradual. STEP 3 Here it is given that g(1) = g(1) = 1 and g"(1) = 10 Thus we can understand that a=1 .The linear approximation formula is given by L(x) = f(a)+(a)(xa) L(x) = 1+1(x1) = 1+x1 = x Thus we have L(x) = x STEP 2 The graph will be Here the black line is the actual curve and the other line is the linear approximation.since g"(1) = 10 is very large,the change in slope will be more gradual. STEP 3 (a). In both the above graphs we can find that the nature of the function differs with the value of 2nd derivative ,but the linear approximation is same. Near x=1 g"(x) > f"(x) Therefore the rate of change of slope is much slower in f than g.So f will have more accurate linear approximation value than g. STEP 3 (b). From the above graphs it is clear that the value of |f"(a)|is very small and the slope is less gradual.and the linear approximation is more accurate. Therefore the error will increase with increasing value of |f"(a)| near x=a. So, the error is proportional to the magnitude of f"(a).