Question

a. graph ( f(x) = \begin{cases} 4 - x^2, & x
eq -2 \\ 7, & x = -2 end{cases} )
b. find ( limlimits_{x \to -2^-} f(x) ) and ( limlimits_{x \to -2^+} f(x) ).
c. does ( limlimits_{x \to -2} f(x) ) exist? if so, what is it? if not, why not?
a. choose the correct graph below.
\\( \bigcirc \\) a. \\( \bigcirc \\) b. \\( \bigcirc \\) c. \\( \bigcirc \\) d. (with corresponding graphs shown)

Explanation:

Part a:

To determine the correct graph, we analyze the function \( f(x)=

$$\begin{cases}4 - x^2, & x eq - 2\\7, & x = - 2\end{cases}$$

\)

• The function \(y = 4-x^{2}\) is a downward - opening parabola with vertex at \((0,4)\) (since for a quadratic function \(y = ax^{2}+bx + c\), the vertex is at \(x=-\frac{b}{2a}\), here \(a=- 1,b = 0,c = 4\), so \(x = 0\) and \(y=4\)).
• For \(x

eq - 2\), the graph of \(y = 4 - x^{2}\) is the parabola. At \(x=-2\), the value of \(y = 4-(-2)^{2}=4 - 4=0\) for the parabola part, but the function is defined as \(f(-2)=7\). So at \(x = - 2\), there is a closed dot (a solid point) at \(y = 7\) and an open dot (a hollow point) at \(y = 0\) on the parabola.

Looking at the options:

• Option A: Has a closed dot at \(x=-2,y = 7\) and an open dot at \(x=-2,y = 0\) on the parabola \(y = 4 - x^{2}\). The parabola \(y=4 - x^{2}\) is drawn correctly (vertex at \((0,4)\), passing through \((2,0)\) and \((- 2,0)\) for the parabola part).
• Option B: At \(x=-2\), the closed dot should be at \(y = 7\), but in this option, the closed dot seems to be at the wrong \(y\) - value.
• Option C: At \(x=-2\), the closed dot is missing (it has an open dot at \(y = 7\) which is incorrect, since \(f(-2)=7\) should be a closed dot).
• Option D: At \(x=-2\), the open dot is at the wrong \(y\) - value and the closed dot is also incorrect.

So the correct graph is Option A.

Part b:

To find \(\lim_{x
ightarrow - 2^{-}}f(x)\) and \(\lim_{x
ightarrow - 2^{+}}f(x)\)

Step 1: Find \(\lim_{x

ightarrow - 2^{-}}f(x)\)
As \(x
ightarrow - 2^{-}\) (approaching \(-2\) from the left), \(x
eq - 2\), so we use the function \(f(x)=4 - x^{2}\).

We substitute \(x=-2\) into \(4 - x^{2}\) (since the limit of a polynomial function \(y = 4 - x^{2}\) as \(x
ightarrow a\) is \(4 - a^{2}\)).

\(\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{-}}(4 - x^{2})=4-(-2)^{2}=4 - 4 = 0\)

Step 2: Find \(\lim_{x

ightarrow - 2^{+}}f(x)\)
As \(x
ightarrow - 2^{+}\) (approaching \(-2\) from the right), \(x
eq - 2\), so we use the function \(f(x)=4 - x^{2}\).

\(\lim_{x
ightarrow - 2^{+}}f(x)=\lim_{x
ightarrow - 2^{+}}(4 - x^{2})=4-(-2)^{2}=4 - 4=0\)

Part c:

A limit \(\lim_{x
ightarrow a}f(x)\) exists if and only if \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)\)

We found that \(\lim_{x
ightarrow - 2^{-}}f(x) = 0\) and \(\lim_{x
ightarrow - 2^{+}}f(x)=0\). Since the left - hand limit and the right - hand limit are equal, \(\lim_{x
ightarrow - 2}f(x)\) exists and \(\lim_{x
ightarrow - 2}f(x)=0\)

Final Answers:

a. \(\boldsymbol{\text{Option A}}\)

b. \(\lim_{x
ightarrow - 2^{-}}f(x)=\boldsymbol{0}\), \(\lim_{x
ightarrow - 2^{+}}f(x)=\boldsymbol{0}\)

c. Yes, \(\lim_{x
ightarrow - 2}f(x)\) exists and \(\lim_{x
ightarrow - 2}f(x)=\boldsymbol{0}\) (because \(\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{+}}f(x) = 0\))

Answer:

Part a:

To determine the correct graph, we analyze the function \( f(x)=

$$\begin{cases}4 - x^2, & x eq - 2\\7, & x = - 2\end{cases}$$

\)

• The function \(y = 4-x^{2}\) is a downward - opening parabola with vertex at \((0,4)\) (since for a quadratic function \(y = ax^{2}+bx + c\), the vertex is at \(x=-\frac{b}{2a}\), here \(a=- 1,b = 0,c = 4\), so \(x = 0\) and \(y=4\)).
• For \(x

eq - 2\), the graph of \(y = 4 - x^{2}\) is the parabola. At \(x=-2\), the value of \(y = 4-(-2)^{2}=4 - 4=0\) for the parabola part, but the function is defined as \(f(-2)=7\). So at \(x = - 2\), there is a closed dot (a solid point) at \(y = 7\) and an open dot (a hollow point) at \(y = 0\) on the parabola.

Looking at the options:

• Option A: Has a closed dot at \(x=-2,y = 7\) and an open dot at \(x=-2,y = 0\) on the parabola \(y = 4 - x^{2}\). The parabola \(y=4 - x^{2}\) is drawn correctly (vertex at \((0,4)\), passing through \((2,0)\) and \((- 2,0)\) for the parabola part).
• Option B: At \(x=-2\), the closed dot should be at \(y = 7\), but in this option, the closed dot seems to be at the wrong \(y\) - value.
• Option C: At \(x=-2\), the closed dot is missing (it has an open dot at \(y = 7\) which is incorrect, since \(f(-2)=7\) should be a closed dot).
• Option D: At \(x=-2\), the open dot is at the wrong \(y\) - value and the closed dot is also incorrect.

So the correct graph is Option A.

Part b:

To find \(\lim_{x
ightarrow - 2^{-}}f(x)\) and \(\lim_{x
ightarrow - 2^{+}}f(x)\)

Step 1: Find \(\lim_{x

ightarrow - 2^{-}}f(x)\)
As \(x
ightarrow - 2^{-}\) (approaching \(-2\) from the left), \(x
eq - 2\), so we use the function \(f(x)=4 - x^{2}\).

We substitute \(x=-2\) into \(4 - x^{2}\) (since the limit of a polynomial function \(y = 4 - x^{2}\) as \(x
ightarrow a\) is \(4 - a^{2}\)).

\(\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{-}}(4 - x^{2})=4-(-2)^{2}=4 - 4 = 0\)

Step 2: Find \(\lim_{x

ightarrow - 2^{+}}f(x)\)
As \(x
ightarrow - 2^{+}\) (approaching \(-2\) from the right), \(x
eq - 2\), so we use the function \(f(x)=4 - x^{2}\).

\(\lim_{x
ightarrow - 2^{+}}f(x)=\lim_{x
ightarrow - 2^{+}}(4 - x^{2})=4-(-2)^{2}=4 - 4=0\)

Part c:

A limit \(\lim_{x
ightarrow a}f(x)\) exists if and only if \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)\)

We found that \(\lim_{x
ightarrow - 2^{-}}f(x) = 0\) and \(\lim_{x
ightarrow - 2^{+}}f(x)=0\). Since the left - hand limit and the right - hand limit are equal, \(\lim_{x
ightarrow - 2}f(x)\) exists and \(\lim_{x
ightarrow - 2}f(x)=0\)

Final Answers:

a. \(\boldsymbol{\text{Option A}}\)

b. \(\lim_{x
ightarrow - 2^{-}}f(x)=\boldsymbol{0}\), \(\lim_{x
ightarrow - 2^{+}}f(x)=\boldsymbol{0}\)

c. Yes, \(\lim_{x
ightarrow - 2}f(x)\) exists and \(\lim_{x
ightarrow - 2}f(x)=\boldsymbol{0}\) (because \(\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{+}}f(x) = 0\))