3 fixed points proof The problem: Suppose $g(x)$ is continously differentiable and has exactly 3 distinct fixed points $r_1 < r_2 < r_3$ with $\lvert g \prime (r_1) \rvert = .5$ and $\lvert g \prime (r_3) \rvert = .5$. Prove that  $ g \prime (r_2) \gt 1$.
My answer:
Let $f(x) = g(x) - x$. Therefore, $f \prime(x)$ is negative at $r_3$ and $r_1$. Since $f \prime(x)$ is negative at $r_3$ and $r_1$, since f only has the 3 0's specified above, and since f is continuous, $f(x) < 0$ for  $x \in (r_1, r_2)$ and $f(x) > 0$ for $x \in (r_2, r_3)$. Thus, since f is differentiable,$f \prime (r_2) >0$.Thus $g \prime (r_2) >1$.
Do I even need continuity of the derivative here? Didn't use it in my proof.
 A: The statement as given is false. Probably not the simplest, but $f(x)=-\frac{x^7}{6}+\frac{x^6}{3}+\frac{x^5}{12}-\frac{2x^4}{3}+\frac{x^3}{12}+\frac{x^2}{3}+x$ has the given derivatives and fixed points ($r_1=-1,r_2=0,r_3=1$), but a slope of $1$ at $r_2$. If you change the $>$ to $\geq$, the statement is true, as proven below:
Suppose the slope at $r_2$ is less than 1. Then for sufficiently small $\epsilon$, $f(r_2-\epsilon)>r_2-\epsilon$. Similarly, the same applies in the neighborhood of $r_1$, such that there is a point $beneath$ $y=x$ in $(r_1,r_2)$. But since $g(x)$ is continuously differentiable, this means that $g$ crosses the line $y=x$ in that interval, contradicting the initial assumption.
A: Let $f(x) = g(x) - x$. Therefore, $f \prime(x)$ is negative at $r_3$ and $r_1$. Since $f \prime(x)$ is negative at $r_3$ and $r_1$, since f only has the 3 0's specified above, and since f is continuous, $f(x) < 0$ for  $x \in (r_1, r_2)$ and $f(x) > 0$ for $x \in (r_2, r_3)$. Thus, since f is differentiable,$f \prime (r_2) >0$.Thus $g \prime (r_2) >1$.
