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)y=f(x) and y=f1(x)y=f^{-1}(x) are symmetrical about the line y=xy=x and intersect on the line y=xy=x.

OR

f(x)=f1(x)=xf(x)=f^{-1}(x)=x whenever graphs intersect.

Now here are some points (questions) :

  1. So if we talk about f(x)=(116)xf(x)=(\frac{1}{16})^x and its inverse functions y=log116xy=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 log116x=(116)xlog_{\frac{1}{16}}x=(\frac{1}{16})^x But algebraically two solutions x=14andx=12x=\frac{1}{4} and x=\frac{1}{2} So now the point is exactly how many roots are there for the equation log116x=(116)xlog_{\frac{1}{16}}x=(\frac{1}{16})^x

  2. Is the above stated property wrong ?

  3. Can there be functions for which f(x)=f1(x)f(x)=f^{-1}(x) has solutions except which lie on the line y=xy=x. If yes, then finite or infinite number ?

  4. Why does the equation log116x=(116)xlog_{\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
5 years 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)f(x) and f1(x)f^{-1}(x) intersect on the line y=xy=x, but that it. It doesn't say f(x)f(x) and f1(x)f^{-1}(x) ONLY intersect on line y=xy=x.

f(x), f1(x)f(x),~f^{-1}(x) and y=xy=x intersect at x0.3642x\approx0.3642

So f(x)=(116)xf(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 - 5 years 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     f1(x)=x     f(x)=f1(x)=xf(x)=x \space \implies f^{-1}(x)=x \space \implies f(x)=f^{-1}(x)=x. It is not saying that every f(x)f(x) and f1(x)f^{-1}(x) intersect on the line y=xy=x only. The graphs of f(x)f(x) and f1(x)f^{-1}(x) may intersect at points not lying on y=xy=x and not necessary that these will cut at y=xy=x.

Sandeep Bhardwaj - 5 years 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 - 5 years 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=xy=x .For instance, inverse of y=xy=-x is the same y=xy=-x. So here f(x)=f1(x)f(x)=f^{-1}(x) has infinitely many solutions i.e.everyRealNumberi.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=xy=x.

I would like to share one example here :

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

f1(x)=f^{-1}(x)={x4,if x[5,6]7x,if x[1,2] \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)y=f(x) and y=f1(x)y=f^{-1}(x) intersect at (32,112)(\frac{3}{2},\frac{11}{2}) and (112,32)(\frac{11}{2},\frac{3}{2}) and Also the intersection points don't lie on the line y=xy=x

Sandeep Bhardwaj - 5 years ago

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Any function that satisfies f2(x)=x f^2(x) = x , will be a good example. This is because we have x=f(y),y=f(x) x = f(y), y = f(x) always, and so there are many many solutions.

Apart from f(x)=x f(x) = x , the next easiest ones to find are f(x)=x f(x) = - x , and then f(x)=x+c f(x) = - x + c .
You might find f(x)=1x f(x) = \frac{1}{x} and f(x)=cxc2+1xc f(x) = \frac{ c x - c^2 + 1 } { x - c } .

Calvin Lin Staff - 5 years 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)f(x)=g(x) the solutions are the values of xx where the graph of f(x)f(x) and g(x)g(x) have intersections. When we have the coincidence that g(x)=f1(x)g(x)=f^{-1}(x) we can be sure that some of the solutions will be the intersection points of y=f(x)y=f(x) with y=xy=x. Like in this problem

Hjalmar Orellana Soto - 4 years, 3 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 - 3 years, 3 months ago

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