a. for the function and point below, find f(a). b. determine an equation of the line tangent to the graph of…

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{3}{\sqrt{x}} = 3x^{-\frac{1}{2}}$.

Step2: Find the derivative using power - rule

The power - rule for differentiation is $\frac{d}{dx}(x^n)=nx^{n - 1}$. For $y = 3x^{-\frac{1}{2}}$, $y^\prime=f^\prime(x)=3\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-\frac{3}{2}x^{-\frac{3}{2}}=-\frac{3}{2x^{\frac{3}{2}}}$.

Step3: Evaluate $f^\prime(a)$ at $a = 4$

Substitute $x = a = 4$ into $f^\prime(x)$. $f^\prime(4)=-\frac{3}{2\times4^{\frac{3}{2}}}=-\frac{3}{2\times8}=-\frac{3}{16}$.

Step4: Find $f(a)$ at $a = 4$

$f(4)=\frac{3}{\sqrt{4}}=\frac{3}{2}$.

Step5: Use point - slope form for tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(a,f(a))=(4,\frac{3}{2})$ and $m = f^\prime(4)=-\frac{3}{16}$. $y-\frac{3}{2}=-\frac{3}{16}(x - 4)$. Expand: $y-\frac{3}{2}=-\frac{3}{16}x+\frac{3}{4}$. $y=-\frac{3}{16}x+\frac{3}{4}+\frac{3}{2}=-\frac{3}{16}x+\frac{3 + 6}{4}=-\frac{3}{16}x+\frac{9}{4}$.

Answer:

a. $f^\prime(4)=-\frac{3}{16}$ b. $y =-\frac{3}{16}x+\frac{9}{4}$