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Use the guidelines of this section to sketch the curve.

$ y = \csc x - 2\sin x $, $ 0 < x < pi $

since $\mathrm{f}^{\prime}$ changes from positive to negative at $\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$Therefore the local maximum is at $\mathrm{f}\left(\frac{\pi}{4}\right)=\mathrm{f}\left(\frac{3 \pi}{4}\right)=0$since $\mathrm{f}^{\prime}$ changes from negative to positive at $\left(\frac{\pi}{2}\right)$Therefore, the local minimum is $\mathrm{f}\left(\frac{\pi}{2}\right)=-1$

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 5

Summary of Curve Sketching

Derivatives

Differentiation

Volume

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

01:11

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06:55

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05:24

Use the guidelines of this…

12:06

07:01

05:21

06:46

04:24

07:31

03:58

04:49

05:05

So, um, we're given this function and, um, this domain is also given from zero to pie. So now if you want a graph, we can look for a few things. Intercept. We have one. If we set, why equals to zero? We have an intercept. Our ex intercepts set pi over four zero and also through over for three pi over four zero. We don't have any whiner substance X. All right and symmetry is odd. And you can check that by seeing um whether it's, for example, plugging and negative one and one into the function. And you see that the answer will be the same value except one of them will be negative and the other will be positive. So it's odd. Assam totes. We have vertical aspen totes, No horizontal. We just have a vertical one. Um, and we'll find that in a second. I'll show you how so. First, let's find the first derivative to see increasing and decreasing, So f prime is negative co tangent of X Coast chickens. Times co C can't X minus to co sign X. And if you set that equal to zero to find critical points, we see we have a critical points at Pi over too. And, um, three pi over too. But three private, too, is not in our domain, so we won't have to worry about that. So at pi over too, we have a critical point. So if we do the first derivative test, let me put pie over too. We see value smaller than pie over two. It's gonna be decreasing, and then greater, it's gonna be increasing. So if it's going from decreased to increase, its gonna be a local minimum point at Pi over, too. And if we do test f of pi over to, we're gonna see, um, into original function, we're going to get the value negative one case, that's the first riveted test. And, um, let's test also are, um, end points on our domain just to see what's going on. And if we do look at, um, the UN 0.0, so zero would be here and pie would be here. So if we do look at those we see that it's gonna be undefined. So we have aspirin totes here and here at our end points. Um, so we're gonna write asking totes X equals zero and X equals pi for our vertical where the slope is undefined and using this information, we can graph so from zero to pipe. So this is zero. This is pie. So we know a zero and pi. We have vertical ass into its and we have a minimum value at pi over to one. So Piper to is gonna be the halfway point. Play over two and then O negative one. Sorry, this is negative One. So pi over to one's is our minimum local minimum value, and we know it's gonna be decreasing close to but never touching the s and tow decrease our minimum point and then starts. A increase is because with our, um, first of a test, we can see increasing and decreasing. And this is our function for our domain.

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