How do you find the tangent equation of #f(x) = (sqrt(x) + 1)/(sqrt(x) + 2)# at a point with x=4?

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Answer 1 · 2015-10-26T19:08:07

The general formula for a tangent at point #(x_o,f(x_o))# is

#y-f(x_o)=f'(x_o)*(x-x_o)#

Now we have that #x_o=4# ,#f(4)=3/4# and

#f'(x)=1/(2*(sqrtx+2)^2*sqrtx)# hence

#f'(4)=1/64#

Hence the tangent is

#y-3/4=1/64*(x-3/4)# or

#y=1/64*x+11/16#