How do you find the linearization at x=1 of #y = 2/x#? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer VNVDVI Apr 2, 2018 #L(x)=4-2x# Explanation: #L(x)=f(a)+f'(a)(x-a)# Here, #f(x)=2/x=2x^-1#, #a=1, f(a)=2.# So, #f'(x)=-2/x^2, f'(1)=-2#, #(x-a)=(x-1)# The linearization is then #L(x)=2-2(x-1)# #L(x)=2-2x+2# #L(x)=4-2x# 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 1888 views around the world You can reuse this answer Creative Commons License