Problem $1$

Find the value of the second non$-$zero term in the Taylor series expansion of $f(x)=\sqrt[\phantom{2}]{x}$ about $x=4.$

$($a$)\frac{1}{4}(x-4)$

$f(a)f^{\text{'}}(a)(x-a)f^{\text{'}\text{'}\text{'}}\frac{(a)}{2!}(x-a{)}^2$

$f(x)=\sqrt[\phantom{2}]{x}=x^{\text{'}\text{'}2}=f(-1)=\sqrt[\phantom{2}]{4}=2$

$($b$)\frac{1}{4}(x4)$

$f^{\text{'}}(x)=,=f^{\text{'}}(-1)=$

$($c$)\frac{1}{2}(x-4)$

$($d$)\frac{1}{2}(x4)$

$($e$)\frac{8}{3}(x-4)$