So while practicing some calculus problems for a test, I felt on a relatively small but hard problem that I couldn’t even know where to start from :

The problem ask you to prove that the equation of the graph Γ\Gamma is the result of fusionning two graphs CgCg and CfC_f Given that Γ\Gamma ‘s equation is as follows :

xآ²yآ²−2xآ²y+xآ²−2x+1=0 xآ²yآ² – 2xآ²y + xآ² – 2x + 1 = 0

And that :

f(x) = \frac{x + \sqrt{2x – 1}}{x} f(x) = \frac{x + \sqrt{2x – 1}}{x}

g(x) = \frac{x – \sqrt{2x – 1}}{x} g(x) = \frac{x – \sqrt{2x – 1}}{x}

And by fusionning two graphs I mean that when two graphs get closer togethr until they form one graph . (I couldn’t find the right term in english, given that I study maths generally in arabic ) .

I really don’t have any idea on how to approach this kind of problem, I’m not looking for raw answers, I’m looking for explanation and I’d appreciate if you could give an example 🙂 .

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Are you sure there is not a typo in ff, gg? Without the factor of 2 outside the square root, ff and gg would be what you get by solving for yy the first equation. In that case the first equation would be a curve in the (x,y)(x,y)-plane which can be written as the superposition of the graphs of ff and gg, with no “fusioning” needed.

– GFR

2 days ago

@GFR Ouupps, my fault for the 2 edited .

– Anis Souames

2 days ago

Good! Is my comment clear or should I elaborate?

– GFR

2 days ago

@GFR So If I understand, I need to solve the 1st equation and I get two solutions, one is f(x) and the other is g(x) ? I’d love if you could elaborate it a bit more in an answer 🙂 .

– Anis Souames

2 days ago

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1 Answer

1

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What the graph of a function f: R \rightarrow Rf: R \rightarrow R really is, is the subset of the plane R^2R^2 given by \{(x,y)\in R^2 :y=f(x))\{(x,y)\in R^2 :y=f(x)), or equivalently \{(x,y)\in R^2 :y-f(x)=0)\{(x,y)\in R^2 :y-f(x)=0). This is not the only way to specify a subset of the plane. For example, I could take a function F:R^2 \rightarrow RF:R^2 \rightarrow R and ask you to plot the subset SS of R^2R^2 given by S=\{ (x,y)\in R^2 : F(x,y)=0\}S=\{ (x,y)\in R^2 : F(x,y)=0\} – this is what you have.

Now not all the equations of the form F(x,y)=0F(x,y)=0 correspond to graphs, but in some cases they do, for example F(x,y)=y-x=0F(x,y)=y-x=0. In your case you nearly get a graph, in fact you get two graphs. I will consider a simpler example which has the same features.

Consider the function F(x,y)= y^2-x^2 F(x,y)= y^2-x^2 . Which subset of the plane corresponds to the points for which F(x,y)=0F(x,y)=0? Well, solving for yy you get y=xy=x and y=-xy=-x, so the set \{(x,y)\in R^2 : y^2-x^2=0\}\{(x,y)\in R^2 : y^2-x^2=0\} is the union of two straight lines through the origin, forming an xx. Note that the “x” figure formed by this set cannot be the graph of a function f:R\rightarrow Rf:R\rightarrow R as two values of yy correspond to each xx, but can be written as the union of two graphs.

Finally, in all the examples I solved for yy as a function of xx, but that is not always the best thing to do: sometimes is more convenient to solve for xx or to do something else.