Question

a. for the following function, find f(a). f(x)=\sqrt{3x + 1}, a = 8 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)=□ (simplify your answer.)

Explanation:

Step1: Rewrite the function

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

Step2: Apply the chain - rule

The chain - rule states that if $y = u^n$ and $u = g(x)$, then $y^\prime=nu^{n - 1}\cdot g^\prime(x)$. Here, $n=\frac{1}{2}$ and $u = 3x+1$, $g(x)=3x + 1$ and $g^\prime(x)=3$. So $f^\prime(x)=\frac{1}{2}(3x + 1)^{-\frac{1}{2}}\cdot3=\frac{3}{2\sqrt{3x + 1}}$.

Step3: Find $f^\prime(a)$ when $a = 8$

Substitute $x = 8$ into $f^\prime(x)$. We have $f^\prime(8)=\frac{3}{2\sqrt{3\times8+1}}=\frac{3}{2\sqrt{25}}=\frac{3}{2\times5}=\frac{3}{10}$.

Step4: Find $f(a)$ when $a = 8$

$f(8)=\sqrt{3\times8 + 1}=\sqrt{25}=5$.

Step5: Find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(a,f(a))=(8,5)$ and $m = f^\prime(8)=\frac{3}{10}$. So $y - 5=\frac{3}{10}(x - 8)$. Expand it: $y-5=\frac{3}{10}x-\frac{12}{5}$. Then $y=\frac{3}{10}x-\frac{12}{5}+5=\frac{3}{10}x+\frac{13}{5}$.

Answer:

a. $f^\prime(8)=\frac{3}{10}$
b. $y=\frac{3}{10}x+\frac{13}{5}$