I am having a really tough time even getting started on this one

[The Student]

Question: Let f be a differentiable function on the interval [0,x]. Prove that there exists a number c ∈ (0,x) such that f(x) = cf′(c) + f(c).
Hint: consider g(x) = xf(x)

The only thing that seems helpful is that the derivative of xf(x) = xf'(x) + f(x) is the same function as cf'(c) + f(c). But I can't get anywhere from there.

 

[pka]

Question: Let f be a differentiable function on the interval [0,x]. Prove that there exists a number c ∈ (0,x) such that f(x) = cf′(c) + f(c).
Hint: consider g(x) = xf(x)

Your problem is very poorly worded. It took me a while to see it.
Worded this way: Let \(\displaystyle f\) be a differentiable function on the interval \(\displaystyle [0.t]\). Prove that there exists a number \(\displaystyle c\in [0,t]\) such that \(\displaystyle f(t)=cf'(c)+f(c)\)

Now use the mean value theorem on \(\displaystyle g(x)=xf(x)\) note that \(\displaystyle g(0)=0~\&~g'(c)=f(c)+cf'(c).\)

 

[The Student]

Your problem is very poorly worded. It took me a while to see it.
Worded this way: Let \(\displaystyle f\) be a differentiable function on the interval \(\displaystyle [0.t]\). Prove that there exists a number \(\displaystyle c\in [0,t]\) such that \(\displaystyle f(t)=cf'(c)+f(c)\)

Now use the mean value theorem on \(\displaystyle g(x)=xf(x)\) note that \(\displaystyle g(0)=0~\&~g'(c)=f(c)+cf'(c).\)

Thanks again - jeez - I just couldn't think of that.