Question

use the function and graph to answer the questions about existence, limits, and continuity. f(x)=\

$$\begin{cases}-x - 3, & -5\\leq x < -4 \\\\x + 5, & -4 < x < -2 \\\\-\frac{3}{2}x, & -2 < x < 0 \\\\0, & 0\\leq x\\leq1 \\\\2x - 2, & 1 < x < 3 \\\\2, & x = 3 \\\\-2x + 10, & 3 < x < 5 \\\\1, & x = 5\\end{cases}$$

does f(-5) exist? no yes

Explanation:

Step1: Identify the interval for x = - 5

The function is defined as $f(x)=-x - 3$ for $-5\leq x<-4$.

Step2: Substitute x = - 5 into the function

Substitute $x=-5$ into $f(x)=-x - 3$. We get $f(-5)=-(-5)-3$.

Step3: Calculate the value

$f(-5)=5 - 3=2$. Since we can find a value for $f(-5)$ using the given piece - wise function, $f(-5)$ exists.

Answer:

Yes