Question

a. for the function and point below, find f(a).
f(x)=\frac{4}{sqrt{x}}, a = \frac{1}{16}
b. determine an equation of the line tangent to the graph of f at (a,f(a)) for the given value of a.
a. f(a)=□

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{4}{\sqrt{x}}$ as $f(x) = 4x^{-\frac{1}{2}}$.

Step2: Find the derivative using the power - rule

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $f(x)=4x^{-\frac{1}{2}}$, $f^\prime(x)=4\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-2x^{-\frac{3}{2}}$.

Step3: Evaluate $f^\prime(a)$ at $a = \frac{1}{16}$

Substitute $x = a=\frac{1}{16}$ into $f^\prime(x)$. $f^\prime(\frac{1}{16})=-2\times(\frac{1}{16})^{-\frac{3}{2}}=-2\times(16)^{\frac{3}{2}}=-2\times(4^2)^{\frac{3}{2}}=-2\times4^3=-2\times64=-128$.

Step4: Find $f(a)$ at $a=\frac{1}{16}$

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

Step5: Use the point - slope form to find the tangent line equation

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope. Here, $x_1=\frac{1}{16}$, $y_1 = 16$ and $m=-128$. So $y - 16=-128(x-\frac{1}{16})$. Expand to get $y-16=-128x + 8$, or $y=-128x+24$.

Answer:

a. $f^\prime(a)=-128$
b. $y=-128x + 24$