How do you find the linearization at a=1 of #f(x) = ln(x)#? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer Jim H Nov 12, 2015 #L=x# Explanation: #L=f(a)+f'(a)(x-a)# #f(x) = lnx# #f(a) = f(1) =0# #f'(x) = 1/x# #f'(a) = f'(1) = 1/1=1# #L=x# 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 9769 views around the world You can reuse this answer Creative Commons License