find the linearization ( l(x) ) at ( x = a ).\n( f(x)=-4 x^{3}-5 x + 1 quad a = 3 )\n( l(x)= )

Explanation:

Step1: Find ( f(a) )

Substitute ( x = a = 3 ) into ( f(x)=-4x^{3}-5x + 1 ). [ \begin{align*} f(3)&=-4\times(3)^{3}-5\times(3)+1\ &=-4\times27-15 + 1\ &=-108-15 + 1\ &=-122 \end{align*} ]

Step2: Find ( f^{\prime}(x) ) and ( f^{\prime}(a) )

Differentiate ( f(x)=-4x^{3}-5x + 1 ) using the power rule ( (x^{n})^\prime=nx^{n - 1} ). ( f^{\prime}(x)=-12x^{2}-5 ) Substitute ( x = 3 ) into ( f^{\prime}(x) ): ( f^{\prime}(3)=-12\times(3)^{2}-5=-12\times9-5=-108 - 5=-113 )

Step3: Use the linearization formula ( L(x)=f(a)+f^{\prime}(a)(x - a) )

Here ( a = 3 ), ( f(a)=-122 ), ( f^{\prime}(a)=-113 ) [ \begin{align*} L(x)&=-122-113(x - 3)\ &=-122-113x+339\ &=-113x + 217 \end{align*} ]

Answer:

( L(x)=-113x + 217 )