Question

find the linearization ( l(x) ) of ( f(x) ) at ( x = a ).

( f(x)=sqrt{x^{2}+11}, a = 5 )

( l(x)= )

Explanation:

Step1: Find \( f(a) \)

Given \( f(x)=\sqrt{x^{2}+11} \) and \( a = 5 \).
Substitute \( x=a = 5 \) into \( f(x) \):
\( f(5)=\sqrt{5^{2}+11}=\sqrt{25 + 11}=\sqrt{36}=6 \)

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

Use the chain - rule. If \( y=\sqrt{u}=u^{\frac{1}{2}} \) where \( u=x^{2}+11 \).
The derivative of \( y \) with respect to \( u \) is \( y_{u}^{\prime}=\frac{1}{2}u^{-\frac{1}{2}} \), and the derivative of \( u \) with respect to \( x \) is \( u_{x}^{\prime}=2x \).
By the chain - rule \( f^{\prime}(x)=\frac{1}{2}(x^{2}+11)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}+11}} \)

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

Substitute \( x = a=5 \) into \( f^{\prime}(x) \):
\( f^{\prime}(5)=\frac{5}{\sqrt{5^{2}+11}}=\frac{5}{6} \)

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

Substitute \( f(5) = 6 \), \( f^{\prime}(5)=\frac{5}{6} \), and \( a = 5 \) into the formula:
\( L(x)=6+\frac{5}{6}(x - 5)=6+\frac{5}{6}x-\frac{25}{6}=\frac{5}{6}x+\frac{36 - 25}{6}=\frac{5}{6}x+\frac{11}{6} \)

Answer:

\( L(x)=\frac{5}{6}x+\frac{11}{6} \)