We have the function $f:\mathbb{R}_{>-1}\rightarrow \mathbb{R}$ with $$f(x)=\log (1+x)-\frac{x}{\sqrt{1+x}}$$ on an interval $[0,b]\subseteq \mathbb{R}$.
Show that there exists a $\xi\in [0,b]$ such that $f(b)=bf'(\xi)$.
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I have done the following :
We can use here the mean value theorem (MVT).
We have that $f$ is continuous on $[0,b]$ as a sum of a logarithmic function and a fraction.
We also have that $f$ is differentiable on $(0,b)$ as a sum of a logarithmic function and a fraction.
From MVT we get then that there is a $\xi\in (0,b)$ such that $$f'(\xi)=\frac{f(b)-f(0)}{b-0}=\frac{f(b)}{b}\Rightarrow f(b)=bf'(\xi)$$
At the statement we want that $\xi\in [0,b]$, i.e. that $\xi$ is contained in a closed interval.
So do we have to check the endpoints seperately?
So do we have to show that $f(b)=bf'(\xi)$ if $\xi=b$ and if $\xi=0$ ?
For that do we have to calculate the derivative of $f$ ?
You're essentially done. Since $(0,b) \subset [0,b]$ and at least one $\xi \in (0,b)$ satisfies the equation, any such $\xi$ is also in the closed interval, $\xi \in [0,b]$.
Remember, "there exists" means that at least one value satisfies the next clause. You showed there's at least one in the smaller set $(0,b)$. If it also happens to be true at $\xi=0$ and/or $\xi=b$, that would be more solutions in the larger set, but that doesn't matter since you already know there's at least one.