For #f(x)=(2x+1)/(x+2) # what is the equation of the tangent line at #x=1#?

1 Answer
Gió
Nov 2, 2015

I found: #y=1/3x+2/3#

Explanation:

First you need to find the slope #m# of the tangent line. This is found deriving your function and calculating it at #x=1#:

#f'(x)=(2(x+2)-(2x+1))/(x+2)^2=(2x+4-2x-1)/(x+2)^2=3/(x+2)^2#

at #x=1#
#f'(3)=3/(1+2)^2=1/3=m#

Now we need the #y# value of the tangence point as well; we find it setting #x=1# into the original function:
#f(1)=(2+1)/(1+2)=3/3=1#

The equation of a line passing through #x_0=1 and y_0=1# with slope #m=1/3# is given as:
#y-y_0=m(x-x_0)#
#y-1=1/3(x-1)#
#y=1/3x-1/3+1#
#y=1/3x+2/3#

Graphically:
enter image source here

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