To determine the points of discontinuity for the given piecewise function
[tex]\[ f(x)=\left\{\begin{array}{ll}
2^{x+2}-1, & -\infty
we need to evaluate the points where the definition of the function changes. These points are:
- [tex]\(x = -2\)[/tex]
- [tex]\(x = 2\)[/tex]
Now, let's analyze the behavior of the function at these points to check for discontinuity.
1. At [tex]\(x = -2\)[/tex]:
- For [tex]\(x < -2\)[/tex], the function is given by [tex]\(2^{x+2}-1\)[/tex].
- For [tex]\(-2 \le x < 2\)[/tex], the function is given by [tex]\(\sqrt{x+2}\)[/tex].
We need to check if [tex]\( \lim_{{x \to -2^-}} f(x) = f(-2) = \lim_{{x \to -2^+}} f(x) \)[/tex]:
[tex]\[ \lim_{{x \to -2^-}} f(x) = \lim_{{x \to -2^-}} (2^{x+2}-1) = 2^0 - 1 = 1 - 1 = 0 \][/tex]
[tex]\[ f(-2) = \sqrt{-2+2} = \sqrt{0} = 0 \][/tex]
[tex]\[ \lim_{{x \to -2^+}} f(x) = \lim_{{x \to -2^+}} \sqrt{x+2} = \sqrt{0} = 0 \][/tex]
Since all these values are equal ([tex]\(0\)[/tex]), the function is continuous at [tex]\(x = -2\)[/tex].
2. At [tex]\(x = 2\)[/tex]:
- For [tex]\(-2 \le x < 2\)[/tex], the function is given by [tex]\(\sqrt{x+2}\)[/tex].
- For [tex]\(x \ge 2\)[/tex], the function is given by [tex]\(\frac{1}{4} x + \frac{3}{2}\)[/tex].
We need to check if [tex]\( \lim_{{x \to 2^-}} f(x) = f(2) = \lim_{{x \to 2^+}} f(x) \)[/tex]:
[tex]\[ \lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} \sqrt{x+2} = \sqrt{4} = 2 \][/tex]
[tex]\[ f(2) = \frac{1}{4} (2) + \frac{3}{2} = \frac{1}{2} + \frac{3}{2} = 2 \][/tex]
[tex]\[ \lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} \left(\frac{1}{4} x + \frac{3}{2}\right) = \frac{1}{4}(2) + \frac{3}{2} = \frac{1}{2} + \frac{3}{2} = 2 \][/tex]
Since all these values are equal ([tex]\(2\)[/tex]), the function is continuous at [tex]\(x = 2\)[/tex].
Therefore, analyzing the points where the piecewise definition changes shows that the function has points of discontinuity at [tex]\(x = -2\)[/tex] and [tex]\(x = 2\)[/tex].
The correct answer is:
[tex]\[ \boxed{x = -2 \text{ and } 2} \][/tex]
