The discussion revolves around finding a function that satisfies the equation 3/2(g(x))^2 = g''(g), where g is suggested to be the inverse of a function f defined by an integral. The problem is situated within the context of calculus, particularly focusing on derivatives and inverse functions.
Several participants have offered guidance on differentiating the functions and applying the chain rule. There is an ongoing exploration of the relationships between the derivatives of f and g, with some participants expressing uncertainty about their calculations and definitions. The discussion is active, with participants iterating on their understanding and clarifying concepts.
Participants note difficulties in aligning their results and understanding the implications of their calculations. There is a recognition of the need for precision in the definitions and arguments of the functions involved, as well as the potential for numerical checks to validate their analytical approaches.
How does g(x) enter the picture? How is it related to f(x)? What does g''(g) mean?monsmatglad said:i want to find the function that satisfies the below. been trying a lot, but all my tactics have proven useless.
f:[0,=>] -> R
f(x)= [0]\int[/x]( 1/(1+t^3)^0.5)
show that 3/2(g(x))^2 = g''(g)
Mark44 said:How does g(x) enter the picture? How is it related to f(x)? What does g''(g) mean?
monsmatglad said:i want to find the function that satisfies the below. been trying a lot, but all my tactics have proven useless.
f:[0,=>] -> R
f(x)= [0]\int[/x]( 1/(1+t^3)^0.5)
show that 3/2(g(x))^2 = g''(g)
f[x_] := NIntegrate[1/Sqrt[1 + t^3],
{t, 0, x}];
mytable = Table[{f[x], x},
{x, 0, 1, 0.01}];
myeqn = Table[x^n, {n, 0, 10}]
myfit = Fit[mytable, myeqn, x]
p1 = Plot[(3/2)*myfit^2, {x, 0, 1},
PlotStyle -> Red]
p2 = Plot[Evaluate[D[myfit, {x, 2}]],
{x, 0, 1}, PlotStyle -> Blue]
Show[{p1, p2}]jackmell said:Ok, I see it. How about you start with just differentiating f(x) and recall that the derivative of the inverse is:
[tex]\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}[/tex]
Now what can you think to do?
monsmatglad said:Hm. I've been trying to diferentiate and integrate g(x), but the numbers don't seem to add up? did they for you? please show me the way=).
monsmatglad said:hm, that's actually what I've been doing, but with no luck. i get that g'(f(x))=(1+x^3)^0.5
and f'(g(x)) is 1/((1+t^3)^0.5)
monsmatglad said:sorry. f'(x) must be 1/(1+x^3)^0.5. and according to the rule which i don't know what is called, the derivative of g(f(x)) is 1/f'(x). this is what i have been using, but with no success. i don't get the answers to fit together like they should.
monsmatglad said:ok. but what do you get when you derivate g(f(x)) in this problem? I see I've been thinking the wrong things, but i really believe i just need to see the correlation, and then i will understand.
monsmatglad said:hm, yes. f(g(x)) = x. f'(g(x))*g'(x)=1. but even though I've been using the wrong definitions, i think what i now get is sort of similar to what I've tried.
monsmatglad said:yes. but this is really where my tactics failed. i get thet g'(x)= (1+x^3)^0.5 and when i integrate and derivate this, the results don't fit like the are supposed to.
monsmatglad said:hm, f'(g(x)) is 1/(1+x^3)
monsmatglad said:hm, i guess this is what made me fail. and i really don't know what f'(g(x)) is.
monsmatglad said:so f'(g(x)) is 1/(1 + g(x)^3 )^(1/2). ?
monsmatglad said:(1+g(x)^3)^0.5 i guess =)
monsmatglad said:(3/2)*g(x)^2*(1+g(x)^3)^-0.5
monsmatglad said:g'(x)*(3/2)*g(x)^2*(1+g(x)^3)^-0.5?