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Use the guidelines of this section to sketch the …

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Problem 69 Hard Difficulty

Use the guidelines of this section to sketch the curve. In guideline D find an equation of the slant asymptote.

$ y = 1 + \dfrac{1}{2}x + e^{-x} $

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Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 5

Summary of Curve Sketching

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Video Transcript

So the domain for this one is X is all real. The intercepts If we set, um, r X values to zero, we see the wise too. So we have one intercept at 02 symmetry. Um, there is no no symmetry. No, it's now odd or even Assam totes. So we do have a slant ass in tow. And if we look at the limit as X approaches infinity of our function, which is one plus one over to X plus e to the power of negative X, we see that we get the limit of one is one. Um, dilemma is expertise. Infinity of half X is infinity. So plus infinity and our last term, um, e to the negative x. We put a plug in the infinity, we get zero in making infinity. So that's saying that there are no horizontal Aston totes, but we do see a slant. Ask himto, um where our limit does not go to zero. So that's anywhere that's not going to zero. So we see one and infinity, Um, that corresponds to the one and 1/2 X. So therefore, why equals, um, 1/2 x plus one. Or you could have said one plus 1/2 X is our slant asking too. All right, next thing we can do is find of prime. So our first derivative is 1/2 minus e to the negative X and sell for a critical points. And we have X equals Ellen of two. All right? And if we do the first derivative test says his f prime and we look att Ellen of two. And by the way, Ellen of two is approximately, um, 0.7. That's just help us graph and, um, f of Ln of two so into our original function. If you plug in that critical point, we're gonna get approximately 1.8 just so we know where to plot this 1.8. All right, So if we look at, um, anything below Ellen of two, it's gonna be decreasing and anything above it's gonna be increasing. Okay, so it's the first derivative test, and now if we wanna use the second derivative test to help us, we see, um, So the second derivative is why double prime? That's gonna equal e to the negative X. And we know that always has to be greater than zero. It has to be positive. So it's gonna be con cave up on our, um, on our whole domain. Khan gave up. So that's the second derivative test. Okay, And now with this information, we can graph, right? So we know we haven't intercepted 02 So let's say this is one this is gonna be too. And we have our since this is decreasing and then increasing, it's gonna be a local minimum value at 0.7 1.8. So let's say this is one. So 10.7 and 1.8, somewhere in here, we're gonna have ah, local minimum value. And we have a Nascimento at why equals 1/2 X plus one. So 1/2 X. So it's going to go up one over two. So this is just a rough sketch of the awesome, too. So it's our slants. Ask himto and our graph is gonna be con Cade. I mean, it's gonna be decreasing than increasing, but it's always gonna be Khan gave up, so it's gonna look something like this

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Calculus: Early Transcendentals

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