3. I'm interested in nding Iirn rs) Where f (2:) : Q :c>4 (E 4 I learned early on in Calc I: . so I tr}r two dierent methods (i) Plot the graph of the function, then read the limit from the graph. (ii) Evaluate at 52 values near 4, and see if the evaluations stabilize near a particular number. 0.252 0.2515 - 2 0.251 on the inter- 17 4 0.2505 val [1,1] in MATLAB, I get the gure shown on the right. When I plot y : 0.25 _ 2 0.2495 - The graph indicates lirn : 0.25. 3.24 2: _ 4 0.249 0.2485 - 0.248 ' ' ' 3.9 3.95 4 4.05 4.1 When I evaluate f at 2? values very close to 4, I get some unexpected results: 5 l4+109 l 4+1010 4+1011l4+1012l4+1013l4+1014\\4+1015l zz?) l 0.25 l 0.25 0.24998 l 0.25 l 0.24779 I 022727 I 0 l (Note that the 2? values on the right are closer to 0). If I believe the table, I would guess that lin1 2 m:~[} 1E 4 of the big jump in f(35') values between :1: = 4 | 1014 and :1: = 4 + 1015. : 0. But I'm suspicious, because I trust the graph over the table: and I suspect that roundoff error is responsible for the unexpected entries in the table. Z 3: 4 l which ones are causing the unexpected f($) values in the table? (Le. which \"danger scenarios\" are present1 and exactly which arithmetic steps cause them here?) (a) (3 points) Among the arithmetic operations that occur when evaluating (b) (2 points) Algebraically simplify the fraction to reduce roundoff error. Show your steps. (Hint: the f 2 can be written as f V4 suggesting that multiplying by the conjugate can help.) (c) (2 points) Now evaluate the simplied form of f(3:) at :1: : 4+ 1015 (in MATLAB); report the value here. Is this Matlab's value the exact value of f (4 + 1015)