We are given that $I=[a,b]$ a compact subinterval of $\mathbb{R}$ and $f:I \to \mathbb{R}$ is a continuous function that's differentiable with $0 < k_0 \le f^{'}(x) \le k_1$ on $(a,b)$. Suppose that $f(a) < c< f(b)$ and $x^*$ is a solution of $f(x) = c$. We zero in on this solution using the iteration:
$$ x_{k+1} = x_{k} - \frac{4(f(x_k)-c)}{5k_1} $$
and letting $e_k = x^* - x_k$ be the error.
We want to show that this error goes to zero and so the iteration converges to the solution.
I noticed that this technique is basically taking the function $f$ and shifting it down by an amount of $c$ which turns the problem into finding a root of $f$ on the given interval. Then this iteration looks almost exactly like Newton's method, with the only difference being that instead of having $\frac{1}{f^{'}(x)}$ in the second term, we have $\frac{4}{5k_1}$. That's about as far as I got in analyzing this problem before getting stuck.
we have $e_{k+1}=x^*-x_{k+1}$
$=e_k+\frac{ 4(f(x_k)-f(x^*)) }{ 5k_1 }$
but
$f(x_k)-f(x^*)=-e_k f'(d)$
with $ d\in (x^*,x_k)$ so
$$e_{k+1}=e_k(1-\frac{4f'(d) }{ 5k_1 })$$
now we use
$k_0\leq f'(d)\leq k_1$
which gives
$-\frac{4}{5}\leq -\frac{4f'(d)}{5k_1}\leq -\frac{4k_0}{5k_1}$
and
$\frac{1}{4}\leq 1-\frac{4f'(d)}{5k_1}\leq 1-\frac{4k_0}{5k_1}<1$ thus
$|e_{k+1}|\leq |e_k| M $ with $M<1$.
so you can conclude using the fact that
$lim_{n \to \infty} M^n=0$.