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You need to remember what the equation of tangent line to the curve, at a point , is, hence:
`f(x) - f(e) = f'(e)(x - e)`
You need to find derivative of function using quotient rule such that:
`f'(x) = ((ln x)'*x - ln x*(x)')/(x^2)`
`f'(x) = (1/x*x - ln x)/(x^2)`
`f'(x) = (1 - ln x)/(x^2)`
You need to substitute e for x in `f'(x)` such that:
`f'(e) = (1 - ln e)/(e^2) =gt f'(e)=0`
If the slope of tangent line to the graph, at the point `(e,f(e))` is zero, then the line is parallel to x axis and it intercepts y axis at `f(x) = f(e) = (ln e)/e = 1/e` .
Hence, the equation of the tangent line to the graph of f(x), at the point `(e,f(e))` is `y=1/e` .
it should have been in the points: square root e and f(square root e)
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