What is the equation of the line normal to $f(x)=lnx-1/x$ at $x=3$ ?

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Answer 1 · 2016-01-08T20:59:06

$f'(x)=1/x-x^-2=1/x-1/x^2=(x-1)/x^2$

now the line equation is:

$(y-y_0)=m(x-x_0)$

where:

$x_0=3$
$y_0=f(x_0)$
$m=-1/(f'(x_0))$ because we looking for the normal line

then:

$f'(3)=(3-1)/9=2/9$
$-1/(f'(x))=-9/2$
$f(x_0)=ln3-1/3$

$y-(ln3-1/3)=-9/2(x-3)$
$y=-9/2x-27/2+(ln3-1/3)$

What is the equation of the line normal to f(x)=lnx-1/x f(x)=lnx-1/x at x=3 x=3?