What is the equation of the line tangent to # f(x)=2/(4 − x)# at # x=p3#?

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Answer 1 · 2018-01-20T17:47:59

#y=2x-4#

Start by finding the derivative:

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

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

Now find #f'(3)# to get the gradient of the tangent:

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

Also find #f(3)# to get a point of intersection:

#f(3) = 2/(4-3) = 2/1=2#

So we have a line with gradient #m=2# and passes through the point: #(3,2)#. Now use #y-b=m(x-a)#

#y-2=2(x-3)->y=2x-6+2#

#y=2x-4#

I am assuming that the question actually says #x=3#