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(a) Use a graph of $ f $ to give a rough estimate of the intervals of concavity and the coordinates of the points of inflection.(b) use a graph of $ f" $ to give better estimates.

$ f(x) = \sin 2x + \sin 4x $, $ 0 \leqslant x \leqslant \pi $

$$f(x)=\sin 2 x+\sin 4 x \Rightarrow f^{\prime}(x)=2 \cos 2 x+4 \cos 4 x \Rightarrow f^{\prime \prime}(x)=-4 \sin 2 x-16 \sin 4 x$$ (a) From the graph of $f,$ it seems that $f$ is $\mathrm{CD}$ on $(0,0.8), \mathrm{CU}$ on $(0.8,1.6), \mathrm{CD}$ on$(1.6,2.3),$ and $\mathrm{CU}$ on $(2.3, \pi) .$ The inflection points appear to be at (0.8,0.7)$(1.6,0),$ and (2.3,-0.7)(b) From the graph of $f^{\prime \prime}$ (and zooming in near the zeros), it seems that $f$ is $\mathrm{CD}$ on$(0,0.85), \mathrm{CU}$ on $(0.85,1.57), \mathrm{CD}$ on $(1.57,2.29),$ and $\mathrm{CU}$ on $(2.29, \pi)$Refined estimates of the inflection points are $(0.85,0.74),(1.57,0),$ and(2.29,-0.74)

04:30

Wen Z.

01:03

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Derivatives

Differentiation

Volume

Missouri State University

Campbell University

Harvey Mudd College

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So we're given this function f of X is equal to sign two X plus sign four X. And we're gonna use a graph of this function to try and estimate where we're concave up. Concave down and the importance of inflection. So I've already graphed this using gizmos and if we look at this graph, you can see from zero to about somewhere a little bit before one, somewhere around here where concave down this hallway and then somewhere around here we start to be concave up and then again, somewhere like around here ish, we're going to be turning back into concave down and then somewhere maybe around here we're concave up again, since we're now concave up and then the last one would maybe be Somewhere around here is three one ish. So the first thing I'm gonna do is we're actually going to just look at the inflection points first, we're gonna write those down and then we'll look at the intervals of con cavity. So the inflection points, the first one is going to be somewhere around here, so X is equal to around .8. Yes. Um The second one is going to be at around here. Somewhere around here. Maybe at one five ISH. So we have .81.5 And then this third one is somewhere down here, maybe two 2.2 ISH. And then so we have 1.8, one at 1.5 and 1.2 point And one at 2.2. And then this last one is probably at 3.1 ISH. So Or three point maybe. Well I guess this is actually pie. So this is the end of our interval. So we're actually going to not include anything past this. So the only three inflection points are going to be at .8 around 1.5 and then around maybe two We'll say 2.2. So I ps inflection points ah X is equal 2.8 Around 1.5 and 22. And now we're going to look at the intervals of con cavity. So we're concave down on this first interval until our Point, our inflection pointed around .8 and then we're concave up from .8 to our second inflection point and then concave down until our third one and then concave up until pie. So from 2.8 where concave down and from one point 5 to around 2.2 or concave down. So From 0 2.8 and from 1.5-2.2 and then we're going to be concave up on the other intervals so from .8 To 1.5 And from 2.2 Two pi since we're only looking at our function on the interval from 0 to Pi. So now for part D. Of part B. Sorry, what we're gonna do is look at a graph of our second derivative. So first thing I'm gonna do is actually find our second derivative. So the first derivative it's equal to, well the derivative of sine is co sign. And then we have to remember to also remember this two X. In here. So renews the chain rule and have co signed two X times two. So we have to co signed two X. And then we're gonna do similar thing over here. We're gonna have four co sign four x. And then to find the second derivative, we're going to use the chain rule again. And the derivative of co sign is negative signs. So we're going to get -4 Since we have to remember this to sign two x -16 Sign of four x. And now we're going to use a graph of this function to estimate a little bit better estimate the intervals of con cavity and inflection points. So if we go back to demos, I have already graphed this on this demos. And if we look at the interval from zero to around pie should be maybe around here, you can see that we have um three different inflection points that are in between our interval of zero to pi and there at around 00.85 ish. So we were close on this first one around pi over two. So A little bit above 1.5 and then 2.29 ISH. So our three inflection points were a little bit Off, but they're all pretty closer within .1 of the actual answer for pretty much all of these. So We can say that we have an inflection point at around .85 pi over two and at 2.3 so Types X is equal 2.85. Hi over two The lesson is two 29 And then our conch it or khan cavity intervals are going to be, if we go back to our graph, I'm gonna be concave down wherever were negative, so are negative from 02.85 And from pi over 2 to 2.29 and then we're gonna be positive on the other intervals. So from 0.85 and from 0 2.85 and from pi over two To 2.29. And then we're going to be concave up On the other intervals. So from .85 two pi over two and from 2.29 two pi. So these are the intervals in which we're concave up in concave down and these are our inflection points.

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