(Section 2.8: Continuity) 2.8.16 Revisiting Example 13 (Verifying the Conclusion of the IVT)Let fx()=x2. According to the IVT, for any value din 0, 4±²³´, there exists a value cin 0, 2±²³´such that fc=d. We try to find such a value for cin terms of d. Let dbe any arbitrary value in 0, 4±²³´. =dand c±0, 2²³´µ¶c2=dand c±0, 2²³´µ¶c=d, a value in 0, 2²³´µObserve: 0±d±4²0±d±2We do not write c=±d, because either d=0, or our value for cwould fall outside of 0, 2±²³´. Verifying the Conclusion of the IVT: What to Write
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