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

1 Answer
Oct 22, 2017

y=((e^6+4)^2/e^6)(x-6)+(1/(e^6+4))

Explanation:

To find the equation of Normal to the line f(x)=1/(e^x+4 we need to find the slope of the normal and the y coordinate at x_1=6

We know the slope of normal is (-1)/(dy/dx) so ,

y=1/(e^x+4)

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

dy/dx=(-e^x)/(e^x+4)^2

So the slope of the normal ( m_n) is (-1)/(dy/dx)=(-1)/((-e^x)/(e^x+4)^2)

Therefore m_n=(e^x+4)^2/e^x

Now to find the y_1 coordinate when x_1=6

We know f(x)=1/(e^x+4)

f(6)=1/(e^6+4)

Therefore the y_1 coordinate is 1/(e^6+4)

We know the equation of a line is

(y-y_1)=m_n(x-x_1)

(y-1/(e^6+4))=((e^6+4)^2/e^6)(x-6)

Therefore y=((e^6+4)^2/e^6)(x-6)+(1/(e^6+4))