Let $F(x)$ be a continuously differentiable function defined on the interval $[a,b]$ such that $F(a)<0$ and $F(b)>0$ and $$ 0<K_1\leq F'(x)\leq K_2\quad (a\leq x\leq b) $$ Find the unique root of equation $F(x)=0$.
The given hint is to use the contraction mapping theorem i.e., if $f(.)$ is a contraction mapping it has a fixed point. Define $f(x)=x-\lambda F(x)$ $$ |f(x)-f(y)|=\Bigg|(x-y)\Big(1-\lambda\frac{F(x)-F(y)}{x-y}\Big)\Bigg|=|x-y|\Big|1-\lambda\frac{F(x)-F(y)}{x-y}\Big| $$ So if I can show $ \Big|1-\lambda\frac{F(x)-F(y)}{x-y}\Big|$ is smaller than $1$, I can say $f(x)=x$. However, I can't proceed from thereon. Many thanks for any help!