Question

#11 (6 pts) We are given the following information about the function f : 2 x+1 f'(x) = - ; f(0) = 1. 1. x²+1 (I) We will construct a continuous piece-wise linear function, g. an approximation off on the interval [0, 4). (Note: gis an approximation off, NOT a linear approximation!) (a) On the interval [0, 1], g is the linear approximation of the function fata=0. Find an expression for g(x), OSX S1. g(x) = , 0sx s 1. (b) (i) On the interval [1, 2], gis a linear function and its graph is a line whose slope is f'(1). Find an expression for g(x), 1sxs 2. (Note, gis continuous at x = 1) g(x) = 15x52. (ii) Why do we think that the function g might be a good approximation of the function f on the interval [1, 2]?

answer all questions or don't respond

Answer 1

of f(a) at r=a Given that f'(x) = levert if(0) = ! 30. J f'(x) = 1 ** dx = sehr am S*** y f(x) = ln (x²+1) + tan(a) ite giren that if(o) a :: 1 = ln (1) + tom'60) + e 7 lec g(x) is a neat approximation g(x) = f(a) + f'(a) (x-2) @ interval [0, 1] g(x) = f(0) + f'(0) (**o) = 1 + 7 (x-o) :- 1 + x . :: Tamada 1+ , osas! 6 internal [1, 2] 8(x) = f() + f'll)(x-1) a ln(2) + taa') + i + (x-1) en (2) + 3 C ) + tornil) +1 = 0-69 3 + 3(x-) + Tom') + . = 0.693 + 3x -.1.5 +1 + 450 = 32 -0807+1+45º = 3x + 0.193+45° :: 12(x) = { x +0:193+450 15x5.2] © internal [2, 3] g(x); $(2) + f'(2) (x-2) = ln(s) + ton (2) +1 + (x-2) = 1.6094 + tañ(2) +1 + x-2 k+ 0.6098 + tom' (2) . 196) = x+0.6094 + tan' (2) 252531

② internal [3,4] g a f(3) + f'(3) (2-3) Elnllo) + tam! (3)+1+fo (-3) Q: 3025 + tori (3) +1 +7* 2013 = 12 + 1.2025 + tai 6) 900) = 72 + 1.09025 + tań (3), 35*34) we get an expression an ex this internal summarizing g(x) on the . @- - we get [0,4]. 96) - I + OSNSI : 3x +0.193+ 45° , 13032 2+0-6094 + tari ) , 23353 Ik +1 2025 + tan' (3), 3<354