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Use the guidelines of this section to sketch the curve.$$y=1 /\left(1+e^{-x}\right)$$

$\left(0, \frac{1}{2}\right)$

01:39

Carson M.

Calculus 1 / AB

Chapter 4

APPLICATIONS OF DIFFERENTIATION

Section 4

Curve Sketching

Derivatives

Differentiation

Applications of the Derivative

Campbell University

Harvey Mudd College

Boston College

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So the domain of this function is X is all real And there is an intercept at 01/2. There is no symmetry. Um and for the ascent oats we have to check that. So limit as X approaches infinity Over function 1/1 plus E to the negative X. Um So as X approaches infinity, we're going to get one. So as X approaches infinity We have an accent two at Y is one. And if it's if we have from negative infinity, it's important to check both sides. So, negative infinity, we're going to get zero. So as X approaches negative infinity, why is going to Because you have a centered at 0? All right. And now we can find the first derivative to test for increasing decreasing intervals. So why prime is equal to E to the power of X over E to the power of X plus one squared. And this is inconclusive. So you when you set it to zero to find critical points inconclusive But we can still look for common cavity. So Y double prime is equal to negative E to the power of x minus one times. Eat to the power of X. It's positively it's the power of X over E to the power of X Plus one and that's all cubed. All right. And now, um when we solve for X, we get zero. So if we look around zero in our first derivative test a second derivative test story. So this is f double prime around zero. We're going to see that it's concave uh and then it's concave down. So concave up to concave down so that this is an inflection point And notice that it's the same as the intercept. So 01/2 is our inflection point. All right, So let's keep that over here. And now we can graph with the information that we got. And the first thing I would do is draw in my asientos. So I have one at one At y equals one. I have an accent too. And I also have one at y equals zero. All right. So now, um I have an intercept at 01 half. So this is my intercept and it's going to be Anything after zero is going to be concave down. Um and it's going to be approaching as X approaches infinity. Y is going to go to this ascent to. So, and as it approaches negative infinity, it's going to approach this as until and we can't give up. So this is our graph

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