What is the equation of the line normal to f(x)=1ex+4 at x=6?

1 Answer
Oct 22, 2017

y=(e6+4)2e6(x6)+(1e6+4)

Explanation:

To find the equation of Normal to the line f(x)=1ex+4 we need to find the slope of the normal and the y coordinate at x1=6

We know the slope of normal is 1dydx so ,

y=1ex+4

Differentiate both sides with respect to x using the chain rule.

dydx=ex(ex+4)2

So the slope of the normal ( mn) is 1dydx=1ex(ex+4)2

Therefore mn=(ex+4)2ex

Now to find the y1 coordinate when x1=6

We know f(x)=1ex+4

f(6)=1e6+4

Therefore the y1 coordinate is 1e6+4

We know the equation of a line is

(yy1)=mn(xx1)

(y1e6+4)=(e6+4)2e6(x6)

Therefore y=(e6+4)2e6(x6)+(1e6+4)