Question

find a formula for the slope m of the tangent line to $f(x)=x^2 + 6x$ at any point $x = a$.
$m = \boxed{2a + 6}$

1. -/ 0.5 points

find the slope m of the tangent line to $f(x)=\sqrt{2x + 1}$ at the point $(4, 3)$.
$m = \boxed{1}$

Explanation:

Step1: Rewrite the function

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

Step2: Apply chain rule for derivative

Let $u=2x+1$, so $f(u)=u^{\frac{1}{2}}$.
$f'(u)=\frac{1}{2}u^{-\frac{1}{2}}$, $u'(x)=2$.
$f'(x)=\frac{1}{2}(2x+1)^{-\frac{1}{2}} \cdot 2 = \frac{1}{\sqrt{2x+1}}$

Step3: Substitute $x=4$ into derivative

$f'(4)=\frac{1}{\sqrt{2(4)+1}} = \frac{1}{\sqrt{9}}$

Step4: Simplify the expression

$\frac{1}{\sqrt{9}} = \frac{1}{3}$

Answer:

$\frac{1}{3}$