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# Something crazy & New to learn !

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 $$y=f(x)$$ and $$y=f^{-1}(x)$$ are symmetrical about the line $$y=x$$ and intersect on the line $$y=x$$.

OR

$$f(x)=f^{-1}(x)=x$$ whenever graphs intersect.

Now here are some points (questions) :

1. So if we talk about $$f(x)=(\frac{1}{16})^x$$ and its inverse functions $$y=log_{\frac{1}{16}}x$$. How many points of intersection will be there ? The above property says that there will be only 1 solution of $$log_{\frac{1}{16}}x=(\frac{1}{16})^x$$ But algebraically two solutions $$x=\frac{1}{4} and x=\frac{1}{2}$$ So now the point is exactly how many roots are there for the equation $$log_{\frac{1}{16}}x=(\frac{1}{16})^x$$

2. Is the above stated property wrong ?

3. Can there be functions for which $$f(x)=f^{-1}(x)$$ has solutions except which lie on the line $$y=x$$. If yes, then finite or infinite number ?

4. Why does the equation $$log_{\frac{1}{16}}x=(\frac{1}{16})^x$$ 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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###### Check out more problems which can solved easily by sketching their graphs, instead of going algebraically. So, try the set : Can you draw its graph ?

Note by Sandeep Bhardwaj
2 years, 9 months ago

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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 $$x\approx0.3642$$

So $$f(x)=\left(\dfrac{1}{16}\right)^x$$ 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. · 2 years, 9 months ago

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 $$f(x)=x \space \implies f^{-1}(x)=x \space \implies f(x)=f^{-1}(x)=x$$. 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 $$y=x$$. · 2 years, 9 months ago

Let's see if we can find some more example of functions having this sort of rebellion nature. @Sandeep Bhardwaj · 2 years, 9 months ago

Any function that satisfies $$f^2(x) = x$$, 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 $$f(x) = \frac{1}{x}$$ and $$f(x) = \frac{ c x - c^2 + 1 } { x - c }$$. Staff · 2 years, 9 months ago

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 $$i.e. every Real Number$$. 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 :

$$f(x)=$$$$\begin{cases} x+4 ,if \space x\in[1,2] \\-x+7, if \space x\in[5,6]\ \end{cases}$$

$$f^{-1}(x)=$$$$\begin{cases} x-4 ,if \space x\in[5,6]\\7-x, if \space x\in[1,2]\ \end{cases}$$

Here graphs of $$y=f(x)$$ and $$y=f^{-1}(x)$$ intersect at $$(\frac{3}{2},\frac{11}{2})$$ and $$(\frac{11}{2},\frac{3}{2})$$ and Also the intersection points don't lie on the line $$y=x$$ · 2 years, 9 months ago

Checkout the th function- (-x^3 +1) it have 4 points of intersection other than that with y=x · 11 months, 3 weeks ago

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 · 1 year, 11 months ago