Calculus question

[roger]

Hello

Could someone help me with this question please ?

A set of curves which all pass the origin, have equations :

y=f1(x)
y=f2(x)
y=f3(x)...... where f ' n(x) = fn-1(x) and f1(x) = x^2

1.) find the expression for fn(x)

2.) find f2(x) and f3(x)

I don't know where to begin, especially the first question .


thanks


roger

[quasar987]

How is the derivative of a function related to the function itself? Through the operation of the integral. So

f_n'(x) =\frac{df_n}{dx} = f_{n-1}(x) \Leftrightarrow ...

[roger]

Could somebody else explain further please ?

[marlon]

Actually quasars tip is totally ok. I would have given the same answer... Just try it...f1 is x² and the derivative of f2 equals f1 = x²...So in order to find f2, just integrate x² with respect to x. You get x³/3...can you move on from here...


regards
marlon

[quasar987]

Assuming you have genuinely been trying to solving the thing for the past 45 minutes, I'll complete the reasoning...

(Assuming f is continuous,)

\Leftrightarrow df_n = f_{n-1}(x)dx \Leftrightarrow \int_0^x df_n = \int_0^x f_{n-1}(x)dx \Leftrightarrow f_n(x) - f_n(0) = \int_0^x f_{n-1}(x)dx \Leftrightarrow f_n(x) - 0 = \int_0^x f_{n-1}(x)dx \Leftrightarrow f_n(x) = \int_0^x f_{n-1}(x)dx

Because if they pass the origin, when x = 0, f = 0.