find the linearization $l(x)$ of the function $f(x)=x^{\frac{1}{2}}$ at $x = 16$. enter equation

Explanation:

Step1: Find the value of the function at $x = 16$

$f(16)=16^{\frac{1}{2}}=\sqrt{16}=4$

Step2: Find the derivative of the function

$f(x)=x^{\frac{1}{2}}$, using the power - rule $(x^n)^\prime=nx^{n - 1}$, we have $f^\prime(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$

Step3: Find the value of the derivative at $x = 16$

$f^\prime(16)=\frac{1}{2\sqrt{16}}=\frac{1}{2\times4}=\frac{1}{8}$

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

Here $a = 16$, so $L(x)=f(16)+f^\prime(16)(x - 16)=4+\frac{1}{8}(x - 16)$ Expand the right - hand side: $L(x)=4+\frac{1}{8}x-2=\frac{1}{8}x + 2$

Answer:

$L(x)=\frac{1}{8}x + 2$