Good morning sir. I have a little ambiguity about the final answer. In the steps preceding, shouldn't there be a direct square root for both t^2s and the square root of 1also 1?
What if we take derivative of both sides to dx…would it not simplify it? And then final find f of x by integrating the right side of equation…would this not work?
@@neevhingrajia3822 Let me rephrase myself. We're not actually substituting 𝑡 = 𝑥, we are just replacing 𝑡 with 𝑥 as the input variable of 𝑓. The reason we can do this is that the range of 𝑥 (i.e., the set of all positive real numbers) is equal to the range of 𝑡 = 1 ∕ 𝑥, so replacing 𝑡 with 𝑥 doesn't change the domain of 𝑓, it is still defined for all positive real numbers.
i didn´t understand the final swich back T, instead of just replacing all T for X, shoudn´t replace it for (1/X)? wich would give us f(t)=x+x[1+(1/x)^2]^(1/2)
Think it like this way @@davidbechor3776 u got the f(t) = 1/t(1+√1+t^2) if u replace t with any real number then u will get a output. now replace the t with x now if u replace the x with any real numbers then u will get the exact same value so here "t" is only a letter it doesn;t matter that much i might make any mistake so please consider the mistake and please provide right answer.
@@davidbechor3776he used t to see what the function does to it. What the function does to z or y or x or t is the same, saying t=x+1 in the beginning doesn't apply to the end. That's why he can replace t with x at the end
That is the first thing I thought. It should just be the reciprocal subbed into the original equation but in his answer he ends up with an extra factor on the left term.
@@mrnogot4251 he simplifies the radical and takes out 1/√t² as a factor then introduces an absolute value 1/|t|. But what if i dont take out a factor, will it be wrong?
if i am not wrong. the x at the end is when x is used to describe a function. the x at the start is an unknown. yes it is confusing, but as far as i know you should just think of the x at the start as just any other variable.@@jairogen90
@@znhaitIn functional notation, f is the name of the function and x the input space of the function. As a whole f(x) is the output an explicit relation in x.A function does not depend on the name f,x,f(x) but on the relation between x and f(x). It is possible to say things like y=y(x) which confuses people but it just means that y depends only on x. It is weird because y is both a variable and the name of a function. I personally think that it is bad nasty notation but it is used a lot so…
I don't understand why the new variable t is even needed. Why shouldn't it be possible to stay with x and just introduce the inverse into the function?
We have 1 ∕ 𝑥 + 1 ∕ 𝑥⋅√(1 + 𝑥²). Multiplication takes precedence over addition, so we can't add 1 ∕ 𝑥 + 1 ∕ 𝑥 = 2 ∕ 𝑥 and then multiply by √(1 + 𝑥²). Instead, what we can do is to factor out 1 ∕ 𝑥, which gives us 1 ∕ 𝑥⋅(1 + √(1 + 𝑥²)). And since multiplying by 1 ∕ 𝑥 is the same as dividing by 𝑥, we can then rewrite that as (1 + √(1 + 𝑥²)) ∕ 𝑥.
If x is allowed to be negative, the expression derived in this video for f(x) won't work. It's given f(1/x)=x+√(1+x²) -- (i) Suppose x is allowed to be negative. let x=-1. (i) gives: f(1/-1)=-1+√[1+(-1)²]=-1+√(1+1)=-1+√2 i.e, f(-1)=-1+√2 -- (ii) In the video, he derives f(x)=(1/x)[1+√(1+x²)] -- (iii) This can be used to evaluate f(-1) Putting x=-1 in (iii), we get: f(-1)=(1/-1)(1+√[1+(-1)²])=-(1+√2)=-1-√2 -- (iv) You can notice that RHSs of (ii) and (iv) are different
You are right. The only problem that arises is when we try to simplify the expression we got for 𝑓(𝑥). The best we can do is basically 𝑓(𝑥) = 1 ∕ 𝑥 + 1 ∕ |𝑥|⋅√(1 + 𝑥²), unless we want to write it as a piecewise function: 𝑥 < 0 ⇒ 𝑓(𝑥) = 1 ∕ 𝑥 + 1 ∕ (−𝑥)⋅√(1 + 𝑥²) = (1 − √(1 + 𝑥²)) ∕ 𝑥 𝑥 > 0 ⇒ 𝑓(𝑥) = 1 ∕ 𝑥 + 1 ∕ 𝑥⋅√(1 + 𝑥²) = (1 + √(1 + 𝑥²)) ∕ 𝑥
√(𝑡²) is the _positive_ square root of 𝑡², so usually we can't say √(𝑡²) = 𝑡 because if 𝑡 < 0 then 𝑡 is the _negative_ square root of 𝑡². The easy way to get around this is to say √(𝑡²) = |𝑡|, because this way the equation holds for negative values of 𝑡 as well. However, in our case the domain is 𝑡 > 0, which means that 𝑡 is not allowed to be negative, and therefore we can write √(𝑡²) = 𝑡.
Anything divided by zero has "infinitely many solutions" in fact you can choose any number to be your answer however a number sequence or function that goes to infinity got many solutions but only one answer
