Question

use this definition to find f(a) at the given number a.
f(x) = \sqrt{6x + 6}, a = 5

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $f(x)=\sqrt{6x + 6}$ and $a = 5$. First, find $f(a+h)$ and $f(a)$.
$f(a + h)=\sqrt{6(a + h)+6}=\sqrt{6a+6h + 6}$, when $a = 5$, $f(5 + h)=\sqrt{30+6h + 6}=\sqrt{36 + 6h}$. And $f(5)=\sqrt{6\times5+6}=\sqrt{30 + 6}=\sqrt{36}=6$.

Step2: Substitute into the derivative - formula

$f^{\prime}(5)=\lim_{h
ightarrow0}\frac{\sqrt{36 + 6h}-6}{h}$.
Rationalize the numerator by multiplying the fraction by $\frac{\sqrt{36 + 6h}+6}{\sqrt{36 + 6h}+6}$.
We get $\lim_{h
ightarrow0}\frac{(\sqrt{36 + 6h}-6)(\sqrt{36 + 6h}+6)}{h(\sqrt{36 + 6h}+6)}$.
Using the difference - of - squares formula $(A - B)(A + B)=A^{2}-B^{2}$, the numerator becomes $(36 + 6h)-36=6h$.
So the limit is $\lim_{h
ightarrow0}\frac{6h}{h(\sqrt{36 + 6h}+6)}$.

Step3: Simplify the limit

Cancel out the $h$ terms in the numerator and denominator: $\lim_{h
ightarrow0}\frac{6}{\sqrt{36 + 6h}+6}$.

Step4: Evaluate the limit

As $h
ightarrow0$, we substitute $h = 0$ into the expression $\frac{6}{\sqrt{36+6h}+6}$.
We have $\frac{6}{\sqrt{36}+6}=\frac{6}{6 + 6}=\frac{6}{12}=\frac{1}{2}$.

Answer:

$\frac{1}{2}$