find the linearization ( l(x) ) at ( x = a ).\n( f(x)=2 x^{3}+3 x + 1 quad a = 2 )\n( l(x)=square )

Explanation:

Step1: Calculate ( f(a) )

Substitute ( x = a = 2 ) into ( f(x)=2x^{3}+3x + 1 ). [ \begin{align*} f(2)&=2\times(2)^{3}+3\times(2)+1\ &=2\times8 + 6+1\ &=16+6 + 1\ &=23 \end{align*} ]

Step2: Find the derivative ( f^{\prime}(x) )

Differentiate ( f(x)=2x^{3}+3x + 1 ) using the power rule ( (x^{n})^\prime=nx^{n - 1} ). ( f^{\prime}(x)=6x^{2}+3 )

Step3: Calculate ( f^{\prime}(a) )

Substitute ( x = a = 2 ) into ( f^{\prime}(x) ). [ \begin{align*} f^{\prime}(2)&=6\times(2)^{2}+3\ &=6\times4+3\ &=24 + 3\ &=27 \end{align*} ]

Step4: Use the linearization formula ( L(x)=f(a)+f^{\prime}(a)(x - a) )

Substitute ( f(2) = 23 ), ( f^{\prime}(2)=27 ), and ( a = 2 ) into the formula. [ \begin{align*} L(x)&=23+27(x - 2)\ &=23+27x-54\ &=27x-31 \end{align*} ]

Answer:

( L(x)=27x - 31 )