Find the linearization L(x) of the function at a. f(x)=x4+6x2,a=−1

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Find the linearization L(x) of the function at a.  f(x)=x4+6x2,a=−1

Ben Owens · Accepted Answer

Step 1 Consider the given function f(x)=x4+6x2.Linearization L(x) at a is given by, L(x)=f(a)+(x−a)f′(a).Clearly, from the given function f(a)=a4+6a2Step 2Differentiate the given function with respect to x.f′(x)=4x3+12xNow, f′(a)=4a3+12aStep 3The linearization at a=−1 is evaluated as follows.L(x)=(a4+6a2)+(x−a)(4a3+12a)=((−1)4+6(−1)2)+(x−(−1))(4(−1)3+12(−1))=7+(x+1)(−16)=7−16x−16=−16x−9Step 4Therefore, the linearization at a=−1isL(x)=−16x−9

lenkiklisg7 · Accepted Answer

Step 1Linearization through tangent line approximation is achieved with equation:L(x)≈f(a)+f′(a)⋅(x−a)The derivative of f(x) is:f(x)=x4+6x2f′(x)=(x4+6x2)′=(x4)′+(6x2)′=4x3+2⋅6x1=4x3+12xAnd with f(a=−1)=4 and f′(a=−1)=−10  our linearized model is:L(x)=(a4+6a2)+(x−a)(4a3+12a)=((−14+6(−1)2)+(x−(−1))(4(−1)3+12(−1))=7−16⋅x−16=−16⋅x−9AnswerL(x)=−16⋅x−9

nick1337 · Accepted Answer

Step 1f(x)=x4+6x2,a=−1Find f(a)f(−1)=(−1)4+6(−1)2=4Find f&#039;f′(x)=4x3+12xFind f&#039;(a)f′(−1)=4(−1)3+12(−1)=−16Step 2Plug everything into the linearization formula.L(x) = f(a) + f&#039;(a)(x-a)= 4 - 16(x-(-1))= 4 - 16x -10= -10x - 6

Find the linearization L(x) of the function at a. f(x)=x^{4}+6x^{2}, a=-1

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