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.
Best Answer
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 $$f'(m)=5m^4-5$$