TITLE: What is the value of of the expression below
QUESTION [0 upvotes]: If $f(0) =0$ and $f''(x)$ exists then what is the value of $$f'(x)-\frac{f(x)}{x}$$
I tried Tailor and Lagrange theorems but to no avail.
clearly $f''(c)=2\frac{f(x)-xf'(0
)}{x^2}$ where $0<c<x$
but how can i proceed further.
How can i find $f'(0)$

REPLY [1 votes]: Expanding $f(t)$ around $x$ gives $f(t)=f(x)+(t-x)f'(x)+\frac{(t-x)^2}2f''(\tau)$, where $\tau$ lies between $t$ and $x$.
For $t=0$ this gives $0=f(0)=f(x)-xf'(x)+\frac{x^2}2f''(\theta x)$ with $\theta \in (0,1)$ and therefore $f'(x)-\frac{f(x)}x=\frac x2f''(\theta x)$.
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