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

Explanation:

Step1: Recall the derivative formula

For $y = \sqrt{u}=(u)^{\frac{1}{2}}$, by the chain - rule $\frac{dy}{dx}=\frac{1}{2\sqrt{u}}\cdot u'$. Here $u = 2x + 4$, so $u'=2$. Then $f'(x)=\frac{2}{2\sqrt{2x + 4}}=\frac{1}{\sqrt{2x + 4}}$.

Step2: Evaluate $f'(a)$ at $a = 6$

Substitute $x=a = 6$ into $f'(x)$. $f'(6)=\frac{1}{\sqrt{2\times6 + 4}}=\frac{1}{\sqrt{12 + 4}}=\frac{1}{\sqrt{16}}=\frac{1}{4}$.

Step3: Find $f(a)$ at $a = 6$

$f(x)=\sqrt{2x + 4}$, when $x = 6$, $f(6)=\sqrt{2\times6+4}=\sqrt{16}=4$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(a,f(a))=(6,4)$ and $m = f'(a)=\frac{1}{4}$. $y-4=\frac{1}{4}(x - 6)$.

Step5: Rearrange the equation

$y-4=\frac{1}{4}x-\frac{6}{4}$, then $y=\frac{1}{4}x-\frac{3}{2}+4=\frac{1}{4}x+\frac{5}{2}$.

Answer:

a. $\frac{1}{4}$ b. $y=\frac{1}{4}x+\frac{5}{2}$