Question

Calculus question plzzzz

Answer 1

Answer:

Step-by-step explanation:

$f(x)=\frac{1}{\sqrt{x} }\\f'(x)=-\frac{1}{2} x^{-\frac{3}{2}}\\\\f'(4)=-\frac{1}{2} 4^{-\frac{3}{2} }=-\frac{1}{2} \frac{1}{4^\frac{3}{2} } \\=-\frac{1}{2} *\frac{1}{8}=-\frac{1}{16} \\when x=4\\y=f(x)=\frac{1}{\sqrt{4} } =\frac{1}{2} \\\\eq.~ of ~tangent ~line~is\\y=-\frac{1}{2} -\frac{1}{16} (x-4)$