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Show that the curve $ y = (1 + x) / (1 + x^2) $ has three points of inflection and they all lie on one straight line.

\[\begin{aligned}y=& \frac{1+x}{1+x^{2}} \Rightarrow y^{\prime}=\frac{\left(1+x^{2}\right)(1)-(1+x)(2 x)}{\left(1+x^{2}\right)^{2}}=\frac{1-2 x-x^{2}}{\left(1+x^{2}\right)^{2}} \Rightarrow \\y^{\prime \prime} &=\frac{\left(1+x^{2}\right)^{2}(-2-2 x)-\left(1-2 x-x^{2}\right) \cdot 2\left(1+x^{2}\right)(2 x)}{\left[\left(1+x^{2}\right)^{2}\right]^{2}}=\frac{2\left(1+x^{2}\right)\left[\left(1+x^{2}\right)(-1-x)-\left(1-2 x-x^{2}\right)(2 x)\right]}{\left(1+x^{2}\right)^{4}} \\&=\frac{2\left(-1-x-x^{2}-x^{3}-2 x+4 x^{2}+2 x^{3}\right)}{\left(1+x^{2}\right)^{3}}=\frac{2\left(x^{3}+3 x^{2}-3 x-1\right)}{\left(1+x^{2}\right)^{3}}=\frac{2(x-1)\left(x^{2}+4 x+1\right)}{\left(1+x^{2}\right)^{3}}\end{aligned}\]So $y^{\prime \prime}=0 \Rightarrow x=1,-2 \pm \sqrt{3}$. Let $a=-2-\sqrt{3}, b=-2+\sqrt{3},$ and $c=1 .$ We can show that $f(a)=\frac{1}{4}(1-\sqrt{3})$$f(b)=\frac{1}{4}(1+\sqrt{3}),$ and $f(c)=1 .$ To show that these three points of inflection lie on one straight line, we'll show that theslopes $m_{a c}$ and $m_{b c}$ are equal\[\begin{array}{l}m_{a c}=\frac{f(c)-f(a)}{c-a}=\frac{1-\frac{1}{4}(1-\sqrt{3})}{1-(-2-\sqrt{3})}=\frac{\frac{3}{4}+\frac{1}{4} \sqrt{3}}{3+\sqrt{3}}=\frac{1}{4} \\m_{b c}=\frac{f(c)-f(b)}{c-b}=\frac{1-\frac{1}{4}(1+\sqrt{3})}{1-(-2+\sqrt{3})}=\frac{\frac{3}{4}-\frac{1}{4} \sqrt{3}}{3-\sqrt{3}}=\frac{1}{4}\end{array}\]

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Derivatives

Differentiation

Volume

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All right, so we're being asked to show that the curve, like with one plus X over one plus X squared, has to, in fact, important. And they live on one straight. So first of all, find the inflection point. We have to find the second derivative. So? So I found the first derivative. I'm doing this so that you conceived here on the right track. So the first derivative comes out to be one minus x squared, much too X all over one plus x squared. And then this whole thing, it's squared. And in the second derivative, they're very variations to this answer. Um and so the different forms. So you can get two x cube, Uh, plus six x squared, minus six sects minus two one excuse express on first AC squared. And this whole thing is cube. And then you can actually simplify this further by pulling out a factor, too. And then two times x squared plus three x squared, minus three x minus one in one plus X squared in the forgiveness pew. And you, Khun, factor this by grouping or you can just Yeah, you contracted this by grouping, and or you could simplify it and then what you get eventually is two times X minus one in times X square. Put your ex plus horn here all over one plus x squared this whole thing this cube and we're looking for where looking forward, this entire thing, that people do it because that's where inflection point a cruise. So we know that the first point will be ethical is one because we'LL get club in one, you get zero And then the second one just probably known we were here. And yet you have to use the quadratic formula because there's no a nice and igniting a whole number factor. So we're going to go ahead and do that on the other side. So when you buy the quadrant for where you going? Negative for plus or minus. The square root of B squared, which is again for squared. And then my n'est porn A C, which is going to times one times one again over to a butt is one so, too, and then some more simplification eventually down to negative to possum minus three. And she put me in a calculator. You get negative three point seven three approximately and negative. Zero point two six eight Respectfully. So now, now that we know are three points, Uh, where the second revenues equals, you weaken. Now evaluate for where scientists occurs for the second for F double rum and see where exactly the inflection points are. So just going to point seventy. I'm not gonna do a point two six, and then it's going to one. Hey, plugging values Western negative point seventy got negative, but we reduced the number to get a card that negative and positive. So we have inflection point occurring at negative one, seven, three, negative country six, eight and one because that's where they scorn from negative to positive here in positive to negative, negative to positive. So we prove that there are three inflection point. But now, with the proved that they all lie on the same line. So that means that the same slope So the way to do that. Well, first of all, I'm gonna find the values at this point, like just on the functions where everyone is equal to one and then effort with three finest, too. That's equal tio three plus one over four. And then after three class two or that you go to negative three, class one over four. Now we have the point. Will you supply the formula for a slow? And Nika was wide too minus y one overact too. Mine is excellent, so we can evaluate it between her from ah, from room three, minus two to one. So between these two and then between the student, the slope of the same. Then we've proven that they lie on the same line. So, uh, so we do three plus one over four, minus one over with three, minus two, minus one. And this gives us one fourth. And then for the next one, we do negative three with one over four, minus three one for And this is going to be negative. Three months too minus three. Close to. And it's also come back to work for And so that's That's how you know. So we proved that the inflection point line, the same life

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