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 \(y=x\).


\(f(x)=f^{-1}(x)=x\) 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 !


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
4 years, 3 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.

Micah Wood - 4 years, 3 months ago

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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 \(f(x)=x \space \implies f^{-1}(x)=x \space \implies f(x)=f^{-1}(x)=x\). 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 \(y=x\).

Sandeep Bhardwaj - 4 years, 3 months ago

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Let's see if we can find some more example of functions having this sort of rebellion nature. @Sandeep Bhardwaj

Sanjeet Raria - 4 years, 3 months ago

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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 \(i.e. every Real Number\). 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 \(y=x\)

Sandeep Bhardwaj - 4 years, 3 months ago

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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 \( f(x) = \frac{ c x - c^2 + 1 } { x - c } \).

Calvin Lin Staff - 4 years, 2 months ago

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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

Hjalmar Orellana Soto - 3 years, 5 months ago

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

rishabh singhal - 2 years, 5 months ago

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