Question

What is the equation of the normal line of `f(x)=1/x-lnx/x` at `x=1`?

Answer 1

The normal line has equation `y = 1/2x + 1/2`.

Explanation:

Let's start by finding the corresponding `y`-coordinate that the function `f(x)` passes through.

`f(1) = 1/1 - ln(1)/1`

`f(1) = 1 - 0/1`

`f(1) = 1`

Now, let's differentiate using the power rule, the quotient rule and the identity `(lnx)' = 1/x`.

`f(x) = x^(-1) - lnx/x`

`f'(x) = -1x^(-2) - (1/x xx x - 1 xx lnx)/x^2`

`f'(x) = -1/x^2 - (1 - lnx)/x^2`

`f'(x) = (-2 + lnx)/x^2`

`f'(x) = (lnx - 2)/x^2`

We now determine the slope of the tangent by inserting our point into the derivative (since this represents the instantaneous rate of change of the function).

`f'(1) = (ln1 - 2)/1^2`

`f'(1) = (-2+ 0)/1`

`f'(1) = -2`

Hence, the slope of the tangent is `-2`. The normal line is always perpendicular to the tangent, so the slope of the normal line is the negative reciprocal of the tangent.

`:.` The slope of the normal line is `1/2`.

We can now determine the equation of the normal line, because we know the slope `(1)`, and the point of contact `(1, 1)`.

`y - y_1 = m(x- x_1)`

`y - 1 = 1/2(x - 1)`

`y - 1 = 1/2x - 1/2`

`y = 1/2x + 1/2`

Hopefully this helps!