How do you find the linearization at a=1 of f(x) = 2 ln(x)? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer Wataru Apr 1, 2017 L(x)=2(x-1) Explanation: f(x)=2ln x Rightarrow f(1)=2ln(1)=0 f'(x)=2/x Rightarrow f'(1)=2/1=2 Linearization L(x) of f at a=1 is: L(x)=f(a)+f'(a)(x-a)=f(1)+f'(1)(x-1)=2(x-1) I hope that this was clear. Answer link Related questions How do you find the linear approximation of (1.999)^4 ? How do you find the linear approximation of a function? How do you find the linear approximation of f(x)=ln(x) at x=1 ? How do you find the tangent line approximation for f(x)=sqrt(1+x) near x=0 ? How do you find the tangent line approximation to f(x)=1/x near x=1 ? How do you find the tangent line approximation to f(x)=cos(x) at x=pi/4 ? How do you find the tangent line approximation to f(x)=e^x near x=0 ? How do you use the tangent line approximation to approximate the value of ln(1003) ? How do you use the tangent line approximation to approximate the value of ln(1.006) ? How do you use the tangent line approximation to approximate the value of ln(1004) ? See all questions in Using the Tangent Line to Approximate Function Values Impact of this question 2347 views around the world You can reuse this answer Creative Commons License