Answer Key
797
161 . False. It is continuous over (−∞, 0) ∪ (0, ∞). 163 . False. Consider f ( x ) = ⎧ ⎩ ⎨ x if x ≠0 4 if x =0 . 165 . False. IVT only says that there is at least one solution; it does not guarantee that there is exactly one. Consider f ( x ) =cos( x ) on [− π , 2 π ]. 167 . False. The IVT does not work in reverse! Consider ( x −1) 2 over the interval [−2, 2].
169 . R = 0.0001519 m 171 . D = 345,826 km 173 . For all values of a , f ( a ) is defined, lim θ → a
f ( θ ) exists, and lim θ → a
f ( θ ) = f ( a ). Therefore, f ( θ ) is continuous
everywhere. 175 . Nowhere
177 . For every ε >0, there exists a δ >0, so that if 0< | t − b | < δ , then | g ( t )− M | < ε 179 . For every ε >0, there exists a δ >0, so that if 0< | x − a | < δ , then | φ ( x )− A | < ε 181 . δ ≤0.25
183 . δ ≤2 185 . δ ≤1 187 . δ <0.3900 189 . Let δ = ε . If 0< | x −3| < ε , then | x +3−6 | = | x −3| < ε . 191 . Let δ = ε 4 . If 0< | x | < ε 4 , then | x 4 | = x 4 < ε . 193 . Let δ = ε 2 . If 5− ε 2 < x <5, then | 5− x | = 5− x < ε . 195 . Let δ = ε /5. If 1− ε /5< x <1, then | f ( x )−3 | =5 x −5< ε . 197 . Let δ = 3 M . If 0< | x +1 | < 3 M , then f ( x ) = 3 ( x +1) 2 > M . 199 . 0.328 cm, ε =8, δ =0.33, a =12, L =144 201 . Answers may vary. 203 . 0 205 . f ( x )− g ( x ) = f ( x )+(−1) g ( x ) 207 . Answers may vary. Review Exercises 209 . False 211 . False. A removable discontinuity is possible.
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