Transcription of MATH2060A Solution to Assignment 1
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MATH2060A Solution to Assignment 1
MATH2060A Solution to Assignment 1 Section Letf:R Rbe defined byf(x) =x2for rationalxandf(x) = 0 for irrationalx. Showthatfis differentiable atx= 0 and findf (0).We claim thatfis differentiable at 0 withf (0) = 0. Consider the difference quotientf(x) f(0)x 0x6= rational, it is equal toxand, whenxis irrational, it is equal to 0. Therefore, f(x) f(0)x 0 0 |x|.For every >0, we take = , then f(x) f(0)x 0 0 |x|< , x6= 0,|x|< .We conclude thatf (0) = (x) g(c)x c=|f(x)| |f(c)|x c= sgn(x c) f(x) f(c)x c ,sincef(c) = +(c) = limx c+sgn(x c) f(x) f(c)x c =|f (c)|.g (c) = limx c sgn(x c) f(x) f(c)x c = |f (c)|.Hencegis differentiable atciffg +(c) =g (c) |f (c)|= |f (c)| f (c) = (a)f(x) =|x|+|x+ 1|= 2x+ 1,forx 01,for 1 x <0 2x 1,forx < 1 Clearly,f (x) = 2,forx >01,for 1< x <0 2,forx < 1 Forx >0,f(x) f(0)x 0=(2x+ 1) 1x 0= 2 f +(0) = 2 Forx <0,f(x) f(0)x 0=1 1x 0= 0 f (0) = 06= 2 =f +(0).
examine the di erentiability of j at 0 using de niton. Indeed, using the fact the cosine function is even, lim h!0 cosjhj cos0 h 0 = lim h!0 cosh 1 h = 0; from which we conclude that jis also di erentiable at x= 0. Hence jis di erentiable everywhere. A shortcut is to realize that the cosine is an even function, so j(x) = cosxis dif-ferentiable ...
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