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Problem 55 Hard Difficulty
(a) Find an equation of the tangent line to the curve $ y = 2/(1 + e^{-x}) $ at the point (0, 1).
(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.
Answer
a. $y=\frac{1}{2} x+1$
b. $y=x / 2+1$
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Video Transcript
for this problem were given the function y equals 2/1 plus e to the negative X, and we want to find the equation of the tangent line at the 0.1 So remember to get the slope of the tangent line, you find the derivative. So we're going to use the quotient rule to find the derivative of the function. So what we have here is the bottom times, the derivative of the top, minus the top times, a derivative of the bottom. And we use the chain rule to find the derivative of E to the negative X. And then that's over the bottom squared. Next, we want to simplify that, and we want to plug in the number zero for X because we're finding the derivative at the 0.1 So a substitute in zero everywhere we have an X and remember that each of the zero is one, so that simplifies things quite a bit. And that gives us negative two times one times negative, 1/2 squared that simplifies to be 2/4. So that's 1/2. And remember, that is the slope of the tangent line. So now that we have the slope and we have the 0.1 We can use our point slope form of the equation of a line. Why minus y one equals M times X minus X one. We can substitute our slope in there and our 0.1 and that will give us the equation of the tangent line, distribute the 1/2 and then add one to both sides and we have y equals 1/2 X plus one. So that takes care of part A and then for part B. What we want to do is illustrate by graphing the curve and the Tanja line on the same screen. So we grab a calculator, we go to y equals and we type in. Our function is why one and we type in our tangent line is why too. All right now we graph and I'm using zoom decimal number four and the blue one is the curve, and the red one is a tangent line. And here we have the cursor at the 10.1 so we can see that that line does appear to be tangent to the curve. At that point
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