How can I find the values of $x$ where the tangent line is horizontal?
I have done the first part of the problem. And to find the value of $x$, I think I'm just supposed to set the derivative equal to zero. But it's not correct. I have done the problem several times and I'm getting $-3$, but that's wrong.
Suppose that $$f(x)=(2x+-2)^{1/5}.$$
(A) Find an equation for the tangent line to the graph of $f(x)$ at $x=2$.
Tangent line: $$y=\frac{x\cdot2^{\frac15}}5+\frac35\cdot2^{\frac15}\qquad\hbox{correct}$$
(B) Find all values of $x$ where the tangent line is horizontal, and enter them as a comma separated list (e.g., $2,-3,6$). If there are none, enter none.
Average of $x$ values: $$-3\qquad\hbox{incorrect}$$
The equation of the perpendicular line will be:
Best Answer
A horizontal tangent line means it's slope is $0$ so what you want here is to find (average of) all $x$ such that $\dfrac{\textrm dy}{\textrm dx}=0$. In this case, you get $$\dfrac25(2x-2)^{-4/5}=0$$ which has no solutions. So the correct answer should be
none.Hope this helps. :)