How do you find the linearization at a=1 of #f(x) =sqrt(x+3) #?

1 Answer
Bill K.
Nov 5, 2015

Use the formula #L(x)=f(a)+f'(a)(x-a)# with #a=1# to get #f(x) approx L(x)=2+(1/4)(x-1)=x/4 + 7/4#.

Explanation:

The linearization of a differentiable function #f# at a point #x=a# is the linear function #L(x)=f(a)+f'(a)(x-a)#, whose graph is the tangent line to the graph of #f# at the point #(a,f(a))#. When #x approx a#, we get the approximation #f(x) approx L(x)#.

For #f(x)=sqrt(x+3)=(x+3)^{1/2}# we get #f'(x)=1/2 * (x+3)^(-1/2}# so that #f(1)=sqrt(4)=2# and #f'(1)=1/2 * 4^{-1/2) = 1/2 * 1/2 = 1/4#. Hence, #f(x) approx L(x)=2+(1/4)(x-1)=x/4 + 7/4#.

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