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 2013 views around the world You can reuse this answer Creative Commons License