CalcNotes0208-page16

CalcNotes0208-page16 - (Section 2.8: Continuity) 2.8.16...

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(Section 2.8: Continuity) 2.8.16 Revisiting Example 13 (Verifying the Conclusion of the IVT) Let fx () = x 2 . According to the IVT, for any value d in 0, 4 ± ² ³ ´ , there exists a value c in 0, 2 ± ² ³ ´ such that fc = d . We try to find such a value for c in terms of d . Let d be any arbitrary value in 0, 4 ± ² ³ ´ . = d and c ± 0, 2 ² ³ ´ µ c 2 = d and c ± 0, 2 ² ³ ´ µ c = d , a value in 0, 2 ² ³ ´ µ Observe: 0 ± d ± 4 ² 0 ± d ± 2 We do not write c d , because either d = 0 , or our value for c would fall outside of 0, 2 ± ² ³ ´ . Verifying the Conclusion of the IVT: What to Write
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