How do you find the linearization of #sqrt(7+2x)# at the point a=0? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer Narad T. Nov 25, 2017 The answer is #=sqrt7(1+1/7x-1/98x^2+1/686x^3+o(x^3))# Explanation: We need #(1+x)^n=1+n/(1!)x+(n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+o(x^3)# Therefore, #sqrt(7+2x)=sqrt7(1+2/7x)^(1/2)# #=sqrt7(1+1/2*2/7x+((1/2)(-1/2))/2*(2/7)^2x^2+((1/2(-1/2)(-3/2)))/(6)*(2/7x)^3+o(x^3))# #=sqrt7(1+1/7x-1/98x^2+1/686x^3+o(x^3))# 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 1521 views around the world You can reuse this answer Creative Commons License