Question

question
find the slope of the tangent line to the graph of $f(x)=sqrt{x + 3}$ at $x = 6$.
provide your answer below:
m$_{tan}=$

Explanation:

Step1: Recall the limit - definition of the derivative

The derivative of a function $y = f(x)$ at $x=a$ is given by $f^\prime(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $f(x)=\sqrt{x + 3}$ and $a = 6$. First, find $f(6)$ and $f(6 + h)$.
$f(6)=\sqrt{6+3}=\sqrt{9}=3$.
$f(6 + h)=\sqrt{(6 + h)+3}=\sqrt{h + 9}$.

Step2: Substitute into the limit - formula

$f^\prime(6)=\lim_{h
ightarrow0}\frac{\sqrt{h + 9}-3}{h}$.
Rationalize the numerator by multiplying the fraction by $\frac{\sqrt{h + 9}+3}{\sqrt{h + 9}+3}$.
We get $\lim_{h
ightarrow0}\frac{(\sqrt{h + 9}-3)(\sqrt{h + 9}+3)}{h(\sqrt{h + 9}+3)}$.
Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(h + 9)-9=h$.
So the limit is $\lim_{h
ightarrow0}\frac{h}{h(\sqrt{h + 9}+3)}$.

Step3: Simplify the limit

Cancel out the $h$ terms in the numerator and denominator (since $h
eq0$ as $h
ightarrow0$).
We have $\lim_{h
ightarrow0}\frac{1}{\sqrt{h + 9}+3}$.

Step4: Evaluate the limit

Substitute $h = 0$ into the expression $\frac{1}{\sqrt{h + 9}+3}$.
We get $\frac{1}{\sqrt{0 + 9}+3}=\frac{1}{3 + 3}=\frac{1}{6}$.

Answer:

$\frac{1}{6}$