the graph of the function $f(x)$ is shown below. state all values of $x$ in the open interval $-9 < x < 9$…

Explanation:

Step1: Define jump discontinuity

A jump discontinuity occurs at $x=a$ if $\lim_{x \to a^-}f(x)$ and $\lim_{x \to a^+}f(x)$ exist but are not equal, with a finite gap between the two one-sided limits.

Step2: Analyze $x=-9$

Left limit does not exist (outside interval), so not valid.

Step3: Analyze $x=-7$

$\lim_{x \to -7^-}f(x) \approx -3$, $\lim_{x \to -7^+}f(x)$ does not connect (gap), but wait—no, check $x=0$: $\lim_{x \to 0^-}f(x) = -2$, $\lim_{x \to 0^+}f(x) = -7$ (finite unequal limits, jump).

Step4: Analyze $x=2$

$\lim_{x \to 2^-}f(x) = -5$, $\lim_{x \to 2^+}f(x) = -4$ (finite unequal limits, jump).

Step5: Verify other points

$x=-5$: limits match (continuous at closed dot), $x=5$: limits match, $x=9$: right limit does not exist (outside interval).

Answer:

$x = 0$, $x = 2$