Question

let $f(x)=sqrt{64 - x}$. compute $f(0)$ using the limit definition
$f(0)=
find an equation of the tangent line at $x = 0$
$y=$
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Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ at $x = a$ is $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $a = 0$ and $f(x)=\sqrt{64 - x}$, so $f(0)=\sqrt{64-0}=8$ and $f(0 + h)=\sqrt{64 - h}$. Then $f^{\prime}(0)=\lim_{h
ightarrow0}\frac{\sqrt{64 - h}-8}{h}$.

Step2: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator $\sqrt{64 - h}+8$. We get $f^{\prime}(0)=\lim_{h
ightarrow0}\frac{(\sqrt{64 - h}-8)(\sqrt{64 - h}+8)}{h(\sqrt{64 - h}+8)}$. Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator is $(64 - h)-64=-h$. So $f^{\prime}(0)=\lim_{h
ightarrow0}\frac{-h}{h(\sqrt{64 - h}+8)}$.

Step3: Simplify the limit

Cancel out the $h$ terms in the numerator and denominator. Then $f^{\prime}(0)=\lim_{h
ightarrow0}\frac{-1}{\sqrt{64 - h}+8}$. Substitute $h = 0$ into the expression, we have $f^{\prime}(0)=-\frac{1}{16}$.

Step4: Find the equation of the tangent line

The equation of a tangent line to the curve $y = f(x)$ at the point $(x_0,y_0)$ is given by $y - y_0=f^{\prime}(x_0)(x - x_0)$. Here, $x_0 = 0$, $y_0=f(0)=8$ and $f^{\prime}(0)=-\frac{1}{16}$. So $y-8=-\frac{1}{16}(x - 0)$, which simplifies to $y=-\frac{1}{16}x + 8$.

Answer:

$f^{\prime}(0)=-\frac{1}{16}$
$y=-\frac{1}{16}x + 8$