Find the equation of tangent to the curve #y=1/(x-3)# at #x=4#?

2 Answers
Nov 4, 2017

Equation of tangent is #x+y=5#

Explanation:

Equation of a tangent to curve #y=f(x)# at #(x_0,f(x_0))# is

#y-f(x_0)=f'(x_0)(x-x_0)#

Here we have #f(x)=1/(x-3)# and are seeking tangent at #x=4# i.e. #x_0=4#.

As #f(x)=1/(x-3)#, #f(4)=1/1=1# and

#f'(x)=(dy)/(dx)=-1/(x-3)^2# and #f'(4)=-1#

Hence, equation of tangent is

#y-1=-1(x-4)#

or #y-1=-x+4#

or #x+y=5#

graph{(x+y-5)(y-1/(x-3))=0 [-6.83, 13.17, -5.04, 4.96]}

Nov 4, 2017

#x+y-5=0#

Explanation:

We should know that an equation of a line can be formed by #(y-y_1)/(x-x_1)=m# , in which #m# is slope of the line and #(x_1,y_1)# is the point of intersection of the curve and the tangent line.

First , we can find the first derivative #dy/dx# which is the slope of the curve.

#y=1/(x-3)#

#dy/dx=d/dx(1/(x-3)) #

#=[(x-3)d/dx(1)-1d/dx(x-3)]/(x-3)^2#

#=-1/(x-3)^2#

Then, we can find the slope of the curve at #x=4#, which is also the slope of tangent line, by subing #x=4# into #dy/dx#.

#dy/dx|_(x=4)=-1/(4-3)^2=-1/1^2=-1#

Here we get #m=-1#

As, we know that the #x# coordinate of the intersection point is #4#, we can sub #x=4# into the equation of the curve to find the respective #y# coordinate.

#y=1/(x-3)=1/(4-3)=1/1=1#

Therefore, the intersection point #=(4,1)#

Now, we have enough information to form the equation of tangent line.

#(y-y_1)/(x-x_1)=m#

Equation of the tangent line at #x=4#:

#(y-1)/(x-4)=-1#

#y-1=-1(x-4)#

#y-1=-x+4#

#x+y-5=0#

Here is the answer. Hope this can help you :)