a. find the derivative function f for the function f. b. determine an equation of the line tangent to the…

Answer:

a. $f'(x)=\frac{1}{2\sqrt{x + 3}}$ b. $y=\frac{1}{4}x+\frac{7}{4}$

Explanation:

Step1: Rewrite the function

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

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = u^n$, then $y'=n\cdot u^{n - 1}\cdot u'$. Here $u=x + 3$, $n=\frac{1}{2}$, and $u'=1$. So $f'(x)=\frac{1}{2}(x + 3)^{\frac{1}{2}-1}\cdot1=\frac{1}{2\sqrt{x + 3}}$

Step3: Find $f(a)$ and $f'(a)$ when $a = 1$

$f(1)=\sqrt{1+3}=2$, $f'(1)=\frac{1}{2\sqrt{1 + 3}}=\frac{1}{4}$

Step4: 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 $(x_1,y_1)=(a,f(a))=(1,2)$ and $m = f'(a)=\frac{1}{4}$. So $y-2=\frac{1}{4}(x - 1)$. Simplifying gives $y=\frac{1}{4}x+\frac{7}{4}$