How do you find the equation of the tangent line to the graph of f(x) at given point if #f(x)= sqrt(x+1))# at (0,1)?

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Answer 1 · 2018-05-15T06:04:33

#y=1/2x +1#

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

We are to find the tangent to #f(x)# at #(0,1)#

First lets confirm that #(0,1)# is a point on #f(x)#

#f(0) = (0+1)^(1/2) -> f(0) = 1#

Hence, #(0,1)# is a point on #f(x)#

Apply power and chain rules to #f(x)#

#f'(x) = 1/2(x+1)^(-1/2) * d/dx (x+1)#

# =1/(2sqrt(x+1)) * 1#

Now, the slope of #f(x)# at #x=0# is #f'(0)#

#f'(0) = 1/(2sqrt1) = 1/2#

The tangent to #f(x)# at #x=0# will have the form #y=mx+c# and from the above we know that #m=1/2#

Thus, #y =1/2x +c# at #(0,1)#

#:. c=1#

Hence, our tangent is : #y=1/2x +1# as we can see from the graphs below.

graph{(y-sqrt(x+1))(y-1/2x-1)=0 [-1.84, 2.487, -0.099, 2.064]}