For the function $f(x) = x^{1/3}$ on the interval $[1,8]$ find the point $(c,f(c))$ guaranteed by the Mean Value Theorem, at which the slope of the tangent line is equal to the slope of the secant line over the entire interval.
(a) Find the slope of the secant line through the endpoints of the graph on $[1, 8]$.
(b) Find the slope of the tangent line at an arbitrary $x$-value $c$.
(c) Set these two expressions equal to find the value of $c$ satisfying the MVT.
(d) Write your answer as an ordered pair twice, first in exact form and then with both coordinates rounded to four decimal places.
I’m not sure if I did this properly but I got the equation for the (tangent?) line is $y= \frac{x}{7} + \frac{6}{7}$ and for part a I got the slope is $m=\frac{1}{7}$. Any assistance is greatly appreciated!!
You got that the secant line is $y=\frac x7 + \frac x6$. From this we can see that the slope of the secant line is $\frac17$.
So, we are asked to find the slope of the tangent line at an arbitrary value $c$. From the power rule, we get the following: $$ \frac d{dx}x^{1/3}=\frac13x^{-2/3}. $$ When they say to find it for an arbitrary value $c$, they just mean to use this instead: $$ \frac13c^{-2/3}. $$
So, that expression is the slope of the tangent line at $c$. We are looking for when the slope of the tangent line is $1/7$, so we want $$ \begin{align} \frac13c^{-2/3} &= \frac17\\ c^{-2/3} &= \frac37\\ c &= \left(\frac37\right)^{-3/2} \end{align}. $$
That's the exact form of the $c$ where the slope of the tangent line is equal to $\frac17$. That's the $x$ coordinate. To get the $y$ coordinate, just plug that into the function. $$ \left(\left(\frac37\right)^{-3/2}\right)^{1/3}=\left(\frac37\right)^{-1/2}. $$
And that should be it. Your ordered pair is $$ \left(\left(\frac37\right)^{-3/2} , \left(\frac37\right)^{-1/2}\right). $$ Now, for the rounding it to four decimal places, just put that into a calculator.