$\int_0^1\frac{(f(x)-1)^2 -4x^2}{x^{3.5}}\,dx$ exists. Calculate $f(0)$ and $f'(0)$

Question

I've tried somehow using Taylor to try and figure this one out.

Unfortunately, I couldn't seem to get a solid answer. Thank you very much for your help!

Let f be a continuos function, $$f:R\rightarrow R$$ $$f(x), \ f'(x) \ , \ f''(x) \ \text{are continuous} $$

and let the integral $$\int_0^1\frac{(f(x)-1)^2 -4x^2}{x^{3.5}}\,dx \space \text{exist and be finite}$$

Find the value of $$f(0)\ \text{and}\ |f'(0)|$$

Answer 1

Let $a=f(0)-1, b=f'(0)$. Then continuity of the second derivative gives $f(x)= 1+a+ bx+O(x^2)$ and hence the integrand is $$\frac{(a+bx+O(x^2))^2-4x^2}{x^{3.5}} = \frac{a^2+2abx+(b^2+a\times\cdots-4)x^2+O(x^3)}{x^{3.5}}$$ A function like $x^c$ is integrable at the origin only if $c>-1$ (check) and hence the first term in the numerator which may be nonzero is the cubic term.

You should be able to figure out what $a,|b|$ are now.