# Show that there is a unique differentiable function satisfying $f''(x)=a, f'(0)=b$, and $f(0)=c$.

Suppose $$a,b,c\in\mathbb{R}$$ and $$f:\mathbb{R}\to\mathbb{R}$$ is differentiable, $$f''(x)=a \; \forall x$$, $$f'(0)=b$$, and $$f(0)=c$$. Find $$f$$ and prove that it is the unique differentiable function with this property.

Obviously, $$f(x)=\frac{a}{2}x^2+bx+c$$. How can I show that this is the unique function with the above property? I was thinking of using Taylor's theorem and showing that the remainder is zero.

Since $$f''(x)=a \; \forall x$$, $$f'''(x)=0$$.

Apply Taylor's Theorem around $$0$$. There exists a $$p$$ between $$x$$ and $$0$$ such that $$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(p)}{3!}x^3=c+bx+\frac{a}{2}x^2$$.

Is this correct? Does this guarantee uniqueness? More generically, is there a one-to-one correspondence between a differentiable function and its Taylor expansion?

Another approach is to assume two functions $$f(x)$$ and $$g(x)$$ satisfying the given conditions and proving that $$g(x)=f(x)$$ for every $$x$$
Note that $g''(x)=f''(x) =a \implies (g-f)''(x)=0$\$
$$\implies (g-f)'(x)=C$$
Plug in $$x=0$$ and we get $$C=0$$, that is $$g'(x)=f'(x)$$ for all $$x$$
That and the condition on $$g(0)=f(0)$$ imply $$g(x)=f(x)$$ for all $$x$$