What is the equation of the tangent line of #f(x)=sqrt(x+1) # at #x=0#?

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Answer 1 · 2018-06-13T00:36:27

# 2y-x-2=0#

We have #f(0)=1#, and

#f^'(x) = 1/(2sqrt(x+1))#

and thus

#f^'(0) = 1/2#

So, the required tangent is a straight line with slope #1/2#, passing through #(0,1)#

It has the equation

#y-1 = 1/2(x-0) implies 2y-x-2=0#