Question

use the graph to answer the question.
describe the continuity of the graphed function. select all that apply.
(1 point)
the function is continuous at ( x = - 4 ).
the function has a removable discontinuity at ( x = - 4 ).
the function has a jump discontinuity at ( x = - 1 ).
the function has an infinite discontinuity at ( x = - 1 ).

Explanation:

Step1: Recall the definition of continuity

A function \(y = f(x)\) is continuous at \(x = a\) if \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)\). A removable discontinuity occurs when \(\lim_{x
ightarrow a}f(x)\) exists but \(f(a)\) is not defined or \(f(a)
eq\lim_{x
ightarrow a}f(x)\). A jump discontinuity occurs when \(\lim_{x
ightarrow a^{-}}f(x)\) and \(\lim_{x
ightarrow a^{+}}f(x)\) both exist but \(\lim_{x
ightarrow a^{-}}f(x)
eq\lim_{x
ightarrow a^{+}}f(x)\). An infinite discontinuity occurs when \(\lim_{x
ightarrow a^{-}}f(x)=\pm\infty\) or \(\lim_{x
ightarrow a^{+}}f(x)=\pm\infty\)

Step2: Analyze \(x = - 4\)

At \(x=-4\), the function has a point (it is defined at \(x = - 4\)) and the left - hand limit and the right - hand limit as \(x\) approaches \(-4\) are equal to the value of the function at \(x=-4\). So the function is continuous at \(x=-4\)

Step3: Analyze \(x=-1\)

At \(x = - 1\), \(\lim_{x
ightarrow - 1^{-}}f(x)\) and \(\lim_{x
ightarrow - 1^{+}}f(x)\) do not exist in the sense of a finite value. The function has a vertical asymptote - like behavior (the open circle with a break in the graph near \(x=-1\) is a mis - representation, but if we consider the standard definitions, when the function approaches \(\pm\infty\) near a point). The function has an infinite discontinuity at \(x=-1\)

Answer:

The function is continuous at \(x = - 4\); The function has an infinite discontinuity at \(x=-1\)