What is the equation of the normal line of $f(x)=(x+5)/(e^x-1)$ at $x=1$ ?

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Answer 1 · 2016-01-17T09:59:02

Normal Line:
$y=(e-1)^2/(5e+1)*x-(e-1)^2/(5e+1)+6/(e-1)$

Given $x=1$
solve $f(1)$ :

$f(1)=6/(e-1)$
The point on the curve $(1, 6/(e-1))$

slope for the normal line $m:$

Use $y-y_1=m*(x-x_1)$

$m=(e-1)^2/(5e+1)$

$y-6/(e-1)=(e-1)^2/(5e+1)*x-(e-1)^2/(5e+1)$

$y=(e-1)^2/(5e+1)*x-(e-1)^2/(5e+1)+6/(e-1)$

What is the equation of the normal line of f(x)=(x+5)/(e^x-1)f(x)=(x+5)/(e^x-1) at x=1x=1?