a. for the following function, find f(a).\n f(x)=\\sqrt{2x + 5}, a = 2\n b. determine an equation of the…

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt{2x + 5}=(2x + 5)^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{2}}$ and $u=2x + 5$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{d}{du}(u^{\frac{1}{2}})=\frac{1}{2}u^{-\frac{1}{2}}$. Second, find $\frac{du}{dx}$: $\frac{d}{dx}(2x + 5)=2$. Then $f^\prime(x)=\frac{1}{2}(2x + 5)^{-\frac{1}{2}}\cdot2=\frac{1}{\sqrt{2x + 5}}$.

Step3: Find $f^\prime(a)$ when $a = 2$

Substitute $x = a=2$ into $f^\prime(x)$. $f^\prime(2)=\frac{1}{\sqrt{2\times2 + 5}}=\frac{1}{\sqrt{4 + 5}}=\frac{1}{3}$.

Step4: Find $f(a)$ when $a = 2$

$f(2)=\sqrt{2\times2+5}=\sqrt{4 + 5}=3$.

Step5: Use the point - slope form for the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(a,f(a))=(2,3)$ and $m = f^\prime(2)=\frac{1}{3}$. So $y-3=\frac{1}{3}(x - 2)$. Rearranging gives $y=\frac{1}{3}x+\frac{7}{3}$.

Answer:

a. $\frac{1}{3}$ b. $y=\frac{1}{3}x+\frac{7}{3}$