How do you find the linearization of #f(x) = x^4 + 5x^2# at a=-1? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer VNVDVI Apr 18, 2018 #L(x)=-6x# Explanation: #L(x)=f(a)+f'(a)(x-a)# We're given #a=-1, f(x)=x^4+5x^2,# so #(x-a)=(x-(-1))=(x+1)# #f(a)=f(-1)=(-1)^4+5(-1)^2=1+5=6# #f'(x)=4x^3+10x# #f'(a)=f'(-1)=4(-1)^3+10(-1)=4-10=-6# Then, #L(x)=6-6(x+1)# #L(x)=6-6x-6# #L(x)=-6x# 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 1877 views around the world You can reuse this answer Creative Commons License