f(cosh(x)) = e^x | Only 1% Can Crack This Functional Equation

Added: 2026-05-14

[00:00:01]Good day, viewers.

[00:00:03]You are welcome.

[00:00:05]Here we have a nice functional equations of f of cosh x equals exponential x.

[00:00:16]And what is f of x?

[00:00:18]So, there are two different things here.

[00:00:20]We have hyperbolic functions and exponential function. And we are asked to find f of x. So, to solve this problem is very, very simple as I'm going to be solving it in the next 5 seconds.

[00:00:34]Let's get into it.

[00:00:36]So, when you consider the solution from here, the domain of this initial function is given as hyperbolic cosh h x. So, the h here it makes it an hyperbolic function.

[00:00:53]And uh hyperbolic functions is of cosh x is given as Suppose we have cosh x is given as exponential of x plus exponential of minus [clears throat] x divided by two.

[00:01:13]This is the equivalent um value of cosh x in hyperbolic function and uh in exponential function.

[00:01:23]>> [snorts] >> Then, by replacing the all of this here, we are going to have f of exponential of x plus exponential minus x divided by two equals exponential of x.

[00:01:39]It is now left to us to find f of x.

[00:01:42]But, these are very, very huge. They are very, very big. We need only x. But, here we have e power x e power minus x divided by two. How can we reduce this to f of x? We are going to apply the the method as the all of e power x plus e power minus x divided by two, we let it to be another letter.

[00:02:04]Suppose I make e power of x plus e power of minus x to be equals 2t.

[00:02:13]So, by letting this one to be equals to 2t, then I need to make x the subject of the formula, but it is now on the power of exponential. What to do is we are going to multiply throughout by exponential x. Multiplying 2 by exponential x, this becomes e power x multiplied by e power x plus e power x multiplied by e power of minus x equals 2t multiplied by e power of x.

[00:02:47]And by this, e power x multiplied by e power x becomes one.

[00:02:52]So, e power x multiplied by e power of x becomes e power of 2x, then plus this cancelled and we have it as one equals 2t e power of x.

[00:03:07]And after having this, let's collect 2t multiplied by e power x to this left hand side as we have e power of 2x minus 2t e power of x plus one equals zero.

[00:03:24]So, this forms a quadratic because the power of this one is two.

[00:03:29]Suppose I let if I let e power of e power of x to be equals p for instance.

[00:03:41]So, this one becomes p squared.

[00:03:44]It becomes p squared minus 2tp plus one equals zero.

[00:03:53]So, this is now looking good as a quadratic because here is the power of two at this p. And this is also p at the base, like at the root of this equation.

[00:04:05]So, they are the one we are going to find, not t. So, this 2t uh is just like the coefficient of p. Let's see how we get the value of p out of this and find the value of x.

[00:04:21]From what we have above, we have p power of two minus 2tp + 1 = 0.

[00:04:32]So, this is what we have.

[00:04:34]Then, by applying the quadratic equations, we have a as one, which is coefficient of p squared.

[00:04:43]And we have b as minus 2t, that is coefficient of p.

[00:04:49]And we have c equals one.

[00:04:52]So, having all these, it is enough for us to apply the quadratic equations as uh we are going to call we call from the standard as p equals -b plus or minus the square root of b squared minus 4ac divided by 2a.

[00:05:17]So, we substitute everything in this formula.

[00:05:21]p equals minus open bracket b is given as minus 2t plus or minus the square root of b here is given as minus 2t squared minus 4 multiplied by a is 1 and c is 1 divided by 2a.

[00:05:45]Then, after having this, we are going to have P equals minus times minus, that is plus. Which means this one becomes plus 2t plus or minus the square root of minus will be affected and changed to plus because of this squared. Then 2 squared is 4 and uh we have t squared minus 4 * 1 * 1, that is 4.

[00:06:10]Everything divided by divided by 2.

[00:06:15]So, here we have P equals 2t plus or minus. By taking out 4 and apply the square root on that, we are going to have 2 multiplied by the square root of t squared minus 1 divided by 2.

[00:06:33]So, next we have P equals 2 is common here. By factoring out 2, we have this as t plus or minus the square root of t squared minus 1 divided by 2.

[00:06:50]And when 2 cancels 2 we are going to get P equals t plus or minus the square root of t squared minus 1.

[00:07:01]This is what we have.

[00:07:03]But initially from cosh x cosh x equals exponential x plus exponential minus x divided by 2.

[00:07:16]The exponential x here is always greater than 0 as the x is greater than or equals 0.

[00:07:25]Here is the domain of it.

[00:07:28]Then since we have known this as we need only the positive value here.

[00:07:36]We don't need anything like negative at this point. So, by replacing this P with exponential X, exponential X is always greater than zero. Therefore, we are going to neglect the negative and go with positive only.

[00:07:53]So, by going with positive only, this shows that the value of our f of t will be equals t plus the square root of t squared minus one.

[00:08:06]And by changing this t to X, we have [snorts] f of x equals x plus the square root of x squared minus one.

[00:08:17]So, where this x is greater than or equals one.

[00:08:21]This is how we solve this problem beautifully. Thanks for watching and don't forget to subscribe to this channel. See you in the next video.

[00:08:28]Never stop learning. Bye-bye.