



Numerade Educator



Problem 78 Hard Difficulty

Show that the curves $ y = e^{-x} $ and $ y = -e^{-x} $ touch the curve $ y = e^{-x} \sin x $ at its inflection points.


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Video Transcript

so sure that the curve widely eaten negative et and like a wild night of eternity just acquitted. Weikel, eat a negative x sign of act. I have a sick inflection point. So we're going to find the derivative of dysfunction and where I'm just second find where the inflection point intact. So say why prime? So when you take the first derivative, you get negative e minus two eggs my neck! Okay, who are you? E minus X coz I next into the second derivative You get it e minus X times sine X minus goes on next minus the annex trying to sign X. And then this simplifies Teo Negative too E minor. And we're looking forward the second zero. But that's where the inflection point curve We know this term right here. Khun articles, you know? So where's Kasan exit duel that people in the direct five or two essentially periodic. We know that however too Eyes pi way know that we do a pie A period of city We'LL get cousin X equals zero at n pi plus pie over too. So if we plug it into this function get why he goes e negative pie or two. Time to sign a piratey, which is just one that equals a to the minus pirate. So we know that if you plug it into wife, what's the minus X? Well, we should get the exact dimensions that we get Y e minus pi over too. So you know that this and this checks out and this is good. Now for negative Eastern minus X it's not you have to do is remember that that there is multiple inflection Point drank this opinion second point every end pie. So we have to find a point where each of minus X or equal will be times negative one. So when the cynical negative one Well, you could do three private T or just simply negative too. So if you put in practical purchase Tio Negative five two and I give this wired vehicle Teo E negative pilot, too. Time Sign our negative pilot here for this thing off. This thing is just negative one. This gives us negative e to the minus pilot, too. And now this is much simpler. So for the y equals negative e student minus actual, we just plug in pie to get negative e on time. Negative pilot, too. So we've just proven God that these two curve Custer inflection point with us now.