I am trying to solve the following exercise in Kolmogorov's real analysis textbook.
Let $F(x)$ be a continuously differentiable function defined on the interval $[a,b]$ such that $F(a) < 0$, $F(b) > 0$, and \begin{align*} 0 < K_1 \leq F'(x) \leq K_2 \; \; \; (a \leq x \leq b). \end{align*} Use Theorem 1 to find the unique root of the equation $F(x) = 0$.
Hint: Introduce the auxiliary function $f(x) = x - \lambda F(x)$, and choose $\lambda$ such that the theorem works for the equivalent equation $f(x) = x$.
Here is a statement of Theorem $1$:
Theorem 1.
Every contraction mapping $A$ defined on a complete metric space $R$ has a unique fixed points.
I would greatly appreciate some help on how to get started. In particular, I am not sure how to use the hint or how the auxiliary equation plays into the problem. (That is, I do not know how to choose the appropriate $\lambda$ or how $f(x)$ relates to $F(x)$.) Any help would be greatly appreciated.
Take $\lambda=\frac{1}{K_2+1}$ and you will have $|f'(x)| \leq 1-\frac{K_1}{K_2+1}<1$
So $f$ is a contraction.
Thus $\exists y \in [a,b]$ unique such that $f(y)=y\Longleftrightarrow y-\frac{1}{K_2+1}F(y)=y$
Thus exists unique $y\in [a,b]$ such that $F(y)=0$