Answer:
[tex]\sqrt[4]{e};\frac1{4e}[/tex]
Step-by-step explanation:
The first derivative of [tex]f(x)[/tex] will give, for each x, the slope of the tangent at that specific x. Let's calculate the derivative, applying the quotient rule.
[tex]D(\frac AB) = \frac{A'B-AB'}{B^2}\\f'(x)= \frac1{x^8}[(\frac1x)x^4-(lnx)(4x^3)]=\frac1{x^8}[x^3(1-4lnx)]=\frac{1-4lnx}{x^5}[/tex]
Now, to find the point with an horizontal tangent (called "stationary points"), we set the first derivative equal to 0. Considering that we're working with [tex]x > 0[/tex] we deal only with the numerator.
[tex]1-4lnx = 0 \rightarrow lnx= \frac14\\x=e^\frac14 =\sqrt[4]{e}[/tex]
At this point we Replace the value we found in the equation to find it's y coordinate
[tex]f(\sqrt[4]e) = \frac{ln\sqrt[4]e}{\sqrt[4]e^4}= \frac{\frac14}{e} = \frac1{4e}[/tex]
