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

Answer:

a. $\frac{7}{10}$ b. $y=\frac{7}{10}x+\frac{29}{10}$

Explanation:

Step1: Find the derivative of $f(x)$

Let $u = 7x + 4$, then $f(x)=\sqrt{u}=u^{\frac{1}{2}}$. Using the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{du}{dx}=7$ and $\frac{df}{du}=\frac{1}{2}u^{-\frac{1}{2}}$. So $f'(x)=\frac{7}{2\sqrt{7x + 4}}$.

Step2: Evaluate $f'(a)$ at $a = 3$

Substitute $x = 3$ into $f'(x)$. $f'(3)=\frac{7}{2\sqrt{7\times3 + 4}}=\frac{7}{2\sqrt{25}}=\frac{7}{10}$.

Step3: Find $f(a)$ at $a = 3$

$f(3)=\sqrt{7\times3 + 4}=\sqrt{25}=5$.

Step4: Find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(3,5)$ and $m = f'(3)=\frac{7}{10}$. $y - 5=\frac{7}{10}(x - 3)$. Expand: $y-5=\frac{7}{10}x-\frac{21}{10}$. $y=\frac{7}{10}x-\frac{21}{10}+5=\frac{7}{10}x+\frac{- 21 + 50}{10}=\frac{7}{10}x+\frac{29}{10}$.