The equation of the tangent line to the curve are [tex]y=x-1 , y=\frac{xe-e^{2}+8 }{e}[/tex].
It is required to find the solution.
A tangent is a line that touches the edge of a curve or circle at one point, but does not cross it. The tangent line to a curve at a point is that straight line that best approximates (or “clings to”) the curve near that point.
Given:
[tex]y=8\frac{lnx}{x} \\\\y'=8(\frac{1/x*x-lnx*1}{x^{2} } \\\\y'=\frac{1-lnx}{x^{2} }[/tex]
Now, we find the slope of the tangent by inserting the point (1,0) into the derivative.
[tex]m_{tangent}=\frac{1-ln1}{1^{2} }\\\\m_{tangent}=1-0/1\\\\m_{tangent} =1[/tex]
We can now find the equation, because we know the slope and a point (1, 0).
[tex]y-y_{1} =m(x-x_{1} )\\[/tex]
y-0=1(x-1)
y=x-1
We can now find the equation, because we know the slope and a point (e, 8/e).
[tex]y-y_{1} =m(x-x_{1} )\\\\y-8/e=1(x-e)\\\\y=\frac{xe-e^{2}+8 }{e}[/tex]
Therefore, the equation of the tangent line to the curve are [tex]y=x-1 , y=\frac{xe-e^{2}+8 }{e}[/tex].
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