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chapter 2 86

# chapter 2 86 - 150 CHAPTERZ PROBLEMS PLUS I 9(\$ h)—9(\$(\$...

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Unformatted text preview: 150 CHAPTERZ PROBLEMS PLUS I _. 9(\$+h)—9(\$)#. (\$+h)f(\$+h)—\$f(\$) 12‘9(\$)‘l‘i‘% h ‘leo h _. mf(\$+h)—:cf(ac) hf(a:+h) _ . f(:c+h)—f(x) . 12% ———h +——h ”211%.? +mf<\$+h> : mf’(:n) + f (m) because f is differentiable and therefore continuous. 13. (a) Put a: = 0 and y = 0 in the equation: f(0 + 0) : f(0) + f(0) + 02 -O + 0 - O2 : f(0) = 2f(0). Subtracting f(0) from each side of this equation gives f(0) = 0. f(0+h) — f(0) [f(0) +f(h) +02h+0h2] — f(0) 0» W = limo —T— 2 m h :an21imM:1 h—>0 h z-vO m 2 2 (C) f’(:c) I 53%) f(a: + h; A m) _}1ﬂ[f(w)+ f(h) -— mhh + wh ] — ftw) = lim W : lim [it—(hi +m2 ——:ch] : 1+av2 h—>0 h h—>0 h 14. We are given that |f(m)| S m2 for all x. In particular, |f(0)| g 0, but |a| Z 0 for all a. The only conclusion is that f(m;:f(0)\_f<m> lf(m)l<w2 lwzl le .mlgf<m>*g<0>glml.gut 0 m |\$| _ l\$| _ |\$| at# f(0) : 0. Now — 0 . . lim (— |m|) : 0 : lim |x| so by the Squeeze Theorem, lim % = 0. So by the deﬁnltion ofa 1H0 xAO zaO derivative. f is differentiable at 0 and, furthermore, f /(0) = 0. ...
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