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#.

Related questions
See all questions in Using the Tangent Line to Approximate Function Values
Impact of this question
52687 views around the world
You can reuse this answer
Creative Commons License