# 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.

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.

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 $$f'(m)=5m^4-5$$

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

$$r=\frac{1}{12}\left(-3-\frac{(15)^{2/3}}{(4\sqrt{6}-9)^{1/3}}+(15(4\sqrt{6}-9)^{1/3}\right)\approx. -0.605829586188268$$

• There are a few subtleties that I have explained herein. Please let me know how I can improve my answer. I really just want to give you the very best answer I can. – Mark Viola Jul 23 '15 at 16:20