Determine the value for $$ c $$ so that $$ \lim _{x ightarrow-2} f(x) $$ exists. $$ f(x)=\left\{\begin{array}{cc}x^{2}-7, & ext { for } x-2\end{array} ight. $$

Question

Determine the value for $$ c $$ so that $$ \lim _{x ightarrow-2} f(x) $$ exists. $$ f(x)=\left\{\begin{array}{cc}x^{2}-7, & 	ext { for } x<-2 \ -x^{2}+c, & 	ext { for } x>-2\end{array}ight. $$

UpStudy AI · Accepted Answer

We need the left-hand limit and the right-hand limit at $$ x = -2 $$ to be equal for the limit to exist.

1. **Left-hand limit:**

For $$ x < -2 $$, we have
   $$
   f(x) = x^2 - 7.
   $$
   Thus, 
   $$
   \lim_{x \to -2^-} f(x) = (-2)^2 - 7 = 4 - 7 = -3.
   $$

2. **Right-hand limit:**

For $$ x > -2 $$, we have
   $$
   f(x) = -x^2 + c.
   $$
   Thus,
   $$
   \lim_{x \to -2^+} f(x) = -(-2)^2 + c = -4 + c.
   $$

3. **Equate the limits:**

Set the left-hand and right-hand limits equal:
   $$
   -3 = -4 + c.
   $$
   Solving for $$ c $$:
   $$
   c = -3 + 4 = 1.
   $$

Thus, the value of $$ c $$ is $$ 1 $$.
To ensure the limit $$ \lim_{x \to -2} f(x) $$ exists, set the left-hand limit equal to the right-hand limit. Calculating both: - **Left-hand limit:** $$ (-2)^2 - 7 = -3 $$ - **Right-hand limit:** $$ -(-2)^2 + c = -4 + c $$ Setting them equal: $$ -3 = -4 + c $$ leads to $$ c = 1 $$. So, $$ c = 1 $$.

Determine the value for so that exists.

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