All the wonder Brilliant users and all my followers, this day I am going to share something really very interesting and may be something new for many of you.
Inspired from a question by @Calvin Lin sir, I have got some discussion points in my mind.
Hope, you all are aware of functions (make mathematics awesome and easy). Here I would like to talk about the function itself and its inverse.
Inverse of a function will exist, if and only if it is one-one(injective) function and onto (surjective) function. Combining one-one and onto together, the function is called Bijective function.
There comes a property of inverse functions :
Graphs of and are symmetrical about the line and intersect on the line .
OR
whenever graphs intersect.
Now here are some points (questions) :
So if we talk about and its inverse functions . How many points of intersection will be there ? The above property says that there will be only 1 solution of But algebraically two solutions So now the point is exactly how many roots are there for the equation
Is the above stated property wrong ?
Can there be functions for which has solutions except which lie on the line . If yes, then finite or infinite number ?
Why does the equation not follow the above property ?
Note: You can add more discussion points here and analyse it as much as you can.
Thank you !
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Comments
Am I missing something? Where in property saying that there will be only 1 solution?
Sure it states that f(x) and f−1(x) intersect on the line y=x, but that it. It doesn't say f(x) and f−1(x) ONLY intersect on line y=x.
f(x), f−1(x) and y=x intersect at x≈0.3642
So f(x)=(161)x follows this property.
Though this property is not always true (see Sandeep Bhardwaj's comment for counterexample), it is true for all real bijective functions.
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What I think is that the property stated above is not true, but this property needs to be modified. And it should be that if f(x)=x ⟹f−1(x)=x ⟹f(x)=f−1(x)=x. It is not saying that every f(x) and f−1(x) intersect on the line y=x only. The graphs of f(x) and f−1(x) may intersect at points not lying on y=x and not necessary that these will cut at y=x.
Let's see if we can find some more example of functions having this sort of rebellion nature. @Sandeep Bhardwaj
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I have a lot of examples contradicting the above property. And one of being is take any line perpendicular to the line y=x .For instance, inverse of y=−x is the same y=−x. So here f(x)=f−1(x) has infinitely many solutions i.e.everyRealNumber. I think the property stated in the note is false.
Also there are many examples against the property, even though if we don't take it perpendicular to the line y=x.
I would like to share one example here :
f(x)={x+4,if x∈[1,2]−x+7,if x∈[5,6]
f−1(x)={x−4,if x∈[5,6]7−x,if x∈[1,2]
Here graphs of y=f(x) and y=f−1(x) intersect at (23,211) and (211,23) and Also the intersection points don't lie on the line y=x
Any function that satisfies f2(x)=x, will be a good example. This is because we have x=f(y),y=f(x) always, and so there are many many solutions.
Apart from f(x)=x, the next easiest ones to find are f(x)=−x, and then f(x)=−x+c.
You might find f(x)=x1 and f(x)=x−ccx−c2+1.
I think the only property we can talk about is that if we have an ecuation we can write as f(x)=g(x) the solutions are the values of x where the graph of f(x) and g(x) have intersections. When we have the coincidence that g(x)=f−1(x) we can be sure that some of the solutions will be the intersection points of y=f(x) with y=x. Like in this problem
Checkout the th function- (-x^3 +1) it have 4 points of intersection other than that with y=x