Question

define f(7) in a way that extends f(s) = (s^3 - 343)/(s^2 - 49) to be continuous at s = 7. f(7) = (type an integer or a simplified fraction.)

Explanation:

Step1: Factor the numerator and denominator

We know that $s^{3}-343=(s - 7)(s^{2}+7s + 49)$ (using $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$ with $a = s$ and $b = 7$) and $s^{2}-49=(s - 7)(s + 7)$ (using $a^{2}-b^{2}=(a - b)(a + b)$ with $a=s$ and $b = 7$). So, $f(s)=\frac{(s - 7)(s^{2}+7s + 49)}{(s - 7)(s + 7)}$.

Step2: Simplify the function

Cancel out the common factor $(s - 7)$ for $s
eq7$. We get $f(s)=\frac{s^{2}+7s + 49}{s + 7}$ for $s
eq7$.

Step3: Find the limit as $s$ approaches 7

$\lim_{s
ightarrow7}\frac{s^{2}+7s + 49}{s + 7}=\frac{7^{2}+7\times7 + 49}{7+7}=\frac{49+49 + 49}{14}=\frac{147}{14}=\frac{21}{2}$.

Answer:

$\frac{21}{2}$