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# Use the guidelines of this section to sketch the curve.$y = (1 + e^x)^{-2}$

## see solution

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So the domain for this function is, um X is all real intercepts. We have one at zero on fourth. There's no symmetry. Um, and let's check for us until it's so we're gonna set, um, the limit to infinity and negative infinity. So the limit as X approaches infinity of our function one plus I need to the power of X to negative to Ah, that's gonna give us zero. So why equals zeroes and us into and negative Infinity is going to give us one. So why equals one is another aspirin too? And now we can look at increasing decreasing intervals. So why prime is negative to e to the power of X over. It's the power of explosive one, and that's all cubed. Set that equal to zero. This is inconclusive. So we can't use this test. All right, um, but why Double prime? We can still look at kong cavity to e to the power of X to e to the power of X minus one over you to the power of experts one that's all to the power of four set that equal to zero and receive critical point at X equals negative. Ln of two, which is approximately negative. 0.693 So if we do the second derivative test so this is F double prime. Uh, negative 0.693 It's gonna be con cave down here con cave up here. So this is an inflection point. All right? Now we can graft with the information we have, so we can graft are Assam toes first, let's say this is one. So we have one. Ask himto that y equals one and we have another one. A y equals zero. We have an intercept at 0 1/4 We have an inflection point at about a negative 0.69 threes. That's about here. If this is, um, this is one it's about this is one. So it would be a somewhere around here. So this is where our inflection point would be the X values of it. So it's going to be Kong cave up here, and it's gonna approach as X approaches infinity. It's gonna approach zero. So it's gonna go like this conclave up, and it's gonna still be con cave up until it hits the inflection point. And then it's gonna switch to Colin Cave down and go to our other s and two. And this is our graph

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