Question

2. Inorder for your phone to calibrate, it's important to have a good idea of how different T and T really are but also to be able to do that on the fly, without too much computation TO do this, we'll use linear approx inution. Specifically, we'll letェ= u2/02, so that and then approximate the function 1/V1-2. (a) Calculate the equation of the tangent line to the function f(z) = atェ= 0. (b) Use the tangent line to approximate the values off(0.5)、f(0.1), and f(0.01) (c) Do you expect your approximations from part (b) to be overestimates or underes mates? Explain. d) Without doing any calculation, which of the approximations from part (b) do you expect to be the most accurate, and which do you expect to be the least accurate? (e) Calculate the exact values of (0.5), (0.1), and f (0.01), rounded to four decimal (f) Recall from class that inear approximation is only a good approximation near the Explain places. Do your answers confirm your expectations from parts (c) and (d)? point of tangency. Why, in real-world terms, is it true that tangency in this example? is near the point of (g) Calculate the value of corresponding to the u and c you found in Problem i. If you were to use your tangent line to approximate the valf () at this value of r, what can you say about the accuracy of the approximation?

Answer 1

f(x)- (a) f(O)-y-1 differentiate bs 1-X 1)-3/2 70)-m (1-0)-3/2, eqn of tangent line is x' (b) f(0.5)? we know f(x) f(%)+ (x-%)f,(%) f(0.5) f(0)+(0.5-0)f (0) 1+0.5× 1.25

f(0.1)? f(0.1) f(0)+ (0.1-0)f (0) f(0.5) 1+(0.1)x1.05 f(0.01)-? f(0.01) f(0)(0.01-0)f(0) 2 (c) These values are underestimates of the real values as we have estimated it thro a tangent line (d) since x=0.01 is close to x=0 hen ce f(0.01) is close to real value