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 2051 views around the world You can reuse this answer Creative Commons License