Tangent line on open interval using the mean value theorem I am having problems understanding how to solve the following question:

Find all points in the open interval $(a, b)$ where the tangent line to $y = f(x)$ is parallel to the secant line joining $(a, f(a))$ and $(b, f(b))$ when $f(x) = x^5 − 5x + 1$ with domain $[−1, 1]$.

I have no problems understanding how to solve this equation at a given point using the mean value theorem, but am having problems understanding how to solve this equation on an open interval.
Link to Desmos Graph of my problem 
My current solution is 
$$
  y= (f'(m_2))(x-m_2)+f(m_2)
$$
where $m_2 = \left(\frac{m+5}{5}\right)^{1/4}$ and $m = \frac{f(a)-f(b)}{a-b}$ which is the slope of the secant line.
Please any help would be much appreciated as I do not have the solution and am unsure of the correct solution.
 A: HINT: we have $$m_s=\frac{f(b)-f(a)}{b-a}=\frac{b^5-a^5-5b+5a}{b-a}$$
the slope of the secant can be simplified to $$a^4+a^3b+a^2b^2+ab^3+b^4-5$$
the slope of the tangent is $$f'(m)=5m^4-5$$
A: The tangent line to $y=f(x)$ at a point $x\in [a,b]$ is given by the derivative $f'(x)$.  We will assume that $a<b$.  For 
$$f=x^5-5x+1 \tag 1$$
the derivative is given by
$$f'(x)=5x^4-5\tag 2$$
We wish to find all points for which 
$$f'(x)=\frac{f(b)-f(a)}{b-a}\tag 3$$
The Mean Value Theorem assures that there exists such a point.  Thus, we substitute $(1)$ and $(2)$ into $(3)$ to obtain
$$\begin{align}
5x^4-5&=\frac{(b^5-a^5)-5(b-a)}{b-a}\\\\
&=b^4+ab^3+a^2b^2+a^3b+a^4-5\\\\
x^4&=\frac{b^4+ab^3+a^2b^2+a^3b+a^4}{5}\\\\
x^4&=\frac{b^4}{5}\left(\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1\right)\tag 4
\end{align}$$
It is easy to show that polynomial $x^4+x^3+x^2+x+1>0$ for all $x$.  Thus, the real solutions are given by 
$$x=\pm|b|\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4} \tag 5$$
We now analyze carefully the solutions as given by $(5)$.

CASE 1: $0<a<b$ or $a<b<0$.
For $0<a<x<b$ or $a<x<b<0$, then  
$$x=b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}$$

CASE 2: $a<0<x<b$.
$$x=b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}$$
occurs when $a/b$ satisfies
$$\frac ab <\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}<1$$
which occurs for $r<\frac ab <0$, where 
$$r=\frac{1}{12}\left(-3-\frac{(15)^{2/3}}{(4\sqrt{6}-9)^{1/3}}+(15(4\sqrt{6}-9)^{1/3}\right)\approx. -0.605829586188268$$
is the negative, real-valued root of the equation $4y^4-y^3-y^2-y-1=0$.

CASE 3: $a<x<0<b$.
$$x=-b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}$$
occurs when $a/b$ satisfies
$$\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}<-\frac ab$$
which occurs for $\frac ab <r$, where 

SUMMARY:
The point $x$, where the slope of the tangent line to the curve $y=x^5-5x+1$ is equal to the slope of the secant line on the interval $[a,b]$, is given by 
$$x=
\begin{cases}
b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}, & ab>0\,\,\text{or}\,\,a<0<b,\,\,\text{and}\,\,\frac{a}{b}>r\\\\
-b\left(\frac{\left(\frac{a}{b}\right)^{4}+\left(\frac{a}{b}\right)^{3}+\left(\frac{a}{b}\right)^{2}+\frac{a}{b}+1}{5}\right)^{1/4}, & a<0<b,\,\,\text{and}\,\,\frac{a}{b}<r
\end{cases}
$$
where 
$$r=\frac{1}{12}\left(-3-\frac{(15)^{2/3}}{(4\sqrt{6}-9)^{1/3}}+(15(4\sqrt{6}-9)^{1/3}\right)\approx. -0.605829586188268$$
