How do you find the linearization at a=0 of # f(x) =sqrt(25-x) #? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer Jim H Aug 28, 2017 #L(x) = f(a)+f'(a)(x-a)# Explanation: Here we have #f(0) = 5#, and #f'(x) = -1/(2sqrt(25-x))#, so #f'(0) = -1/10# Therefore, #L(x) = 5-1/10x# 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 3781 views around the world You can reuse this answer Creative Commons License