# Finding the fixed point

Trying to solve this question, got this answer but have a gut feeling that this might not be the way to do it, by the way this topic is related to fixedpoints

The solution that I came up with

-

The problem is motivated by a fixed point iteration procedure, but solving it is an exercise in inequalities. You are told that $$-1\lt f'(x)\lt 1,$$ and want information about the values of $x$ for which these inequalities hold.
We have $f'(x)=-26x+3$. So we start from the inequality $$-1\lt -26x+3\lt 1.$$ Rewrite this as $$-4\lt -26x\lt -2.$$ Now we divide by $-26$, remembering that dividing by a negative number reverses inequalities. We get $$\frac{2}{26}\lt x\lt \frac{4}{26}.$$ Now we are finished. The initial inequality $-1\lt f'(x)\lt 1$ is equivalent to our last inequality, that is, to the inequality $h\lt x\lt k$ with $k=4/26$. We are asked to compute $k$ to $2$ decimal places.
Remark: The connection with the fixed point iteration problem that gave rise to this is that we are trying to solve $ax^2+bx+c=x$, where $a$, $b$, and $c$ are your specific numbers. Let $f(x)=ax^2+bx+c$. Start with a number $x_0$ such that $2/26\lt x_0\lt 4/26$. Let $x_1=f(x_0)$, $x_2=f(x_1)$, $x_3=f(x_2)$, and so on. Then because our derivative has absolute value $\lt 1$ in our interval $(h,k)$, the $x_n$ converge to a solution of $f(x)=x$. This is a toy problem, since for our particular $f(x)$ we can use the Quadratic Formula to solve the equation $f(x)=x$. However, the same ideas work for functions $f(x)$ for which there is no simple algebraic way to solve $f(x)=x$.