Answer :
To determine the intervals of continuity for the function \( f(x) = \begin{cases}
-e^x & \text{if } x \neq 0 \\
\tan x & \text{if } x = 0
\end{cases} \), we need to analyze its behavior at \( x = 0 \) and elsewhere.
1. Continuity for \( x \neq 0 \):
- For \( x \neq 0 \), the function is defined as \( f(x) = -e^x \). The exponential function \( e^x \) is smooth and continuous everywhere, so \( -e^x \) is also continuous for all \( x \). Hence, \( f(x) \) is continuous on \( (-\infty, 0) \cup (0, \infty) \).
2. Continuity at \( x = 0 \):
- At \( x = 0 \), \( f(x) = \tan(0) \). Since \(\tan(0) = 0\), the function value at \( x = 0 \) is defined and equals \( 0 \).
- To determine the continuity at \( x = 0 \), we need to check the limit of \( f(x) \) as \( x \) approaches \( 0 \) from both sides and compare it with the function value at \( x = 0 \):
[tex]\[ \lim_{x \to 0^+} -e^x = -e^0 = -1 \][/tex]
[tex]\[ \lim_{x \to 0^-} -e^x = -e^0 = -1 \][/tex]
- The limit of \( f(x) \) as \( x \) approaches \( 0 \) from both the left and the right is \(-1\).
- However, the value of the function at \( x = 0 \) is \( f(0) = \tan(0) = 0 \).
Since the limit of \( f(x) \) as \( x \) approaches \( 0 \) does not equal the function value at \( x = 0 \), there is a discontinuity at \( x = 0 \).
3. Type of Discontinuity:
- The limit exists and is finite, but it does not equal the function's value at \( x = 0 \). This indicates that there is a removable discontinuity at \( x = 0 \).
Therefore, we summarize:
- The function \( f \) is continuous on \( (-\infty, 0) \cup (0, \infty) \).
- There is a removable discontinuity at \( x = 0 \).
Hence, the correct answer is:
- The function \( f \) is continuous on \( (-\infty, 0) \cup (0, \infty) \).
- There is a removable discontinuity at [tex]\( x = 0 \)[/tex].
-e^x & \text{if } x \neq 0 \\
\tan x & \text{if } x = 0
\end{cases} \), we need to analyze its behavior at \( x = 0 \) and elsewhere.
1. Continuity for \( x \neq 0 \):
- For \( x \neq 0 \), the function is defined as \( f(x) = -e^x \). The exponential function \( e^x \) is smooth and continuous everywhere, so \( -e^x \) is also continuous for all \( x \). Hence, \( f(x) \) is continuous on \( (-\infty, 0) \cup (0, \infty) \).
2. Continuity at \( x = 0 \):
- At \( x = 0 \), \( f(x) = \tan(0) \). Since \(\tan(0) = 0\), the function value at \( x = 0 \) is defined and equals \( 0 \).
- To determine the continuity at \( x = 0 \), we need to check the limit of \( f(x) \) as \( x \) approaches \( 0 \) from both sides and compare it with the function value at \( x = 0 \):
[tex]\[ \lim_{x \to 0^+} -e^x = -e^0 = -1 \][/tex]
[tex]\[ \lim_{x \to 0^-} -e^x = -e^0 = -1 \][/tex]
- The limit of \( f(x) \) as \( x \) approaches \( 0 \) from both the left and the right is \(-1\).
- However, the value of the function at \( x = 0 \) is \( f(0) = \tan(0) = 0 \).
Since the limit of \( f(x) \) as \( x \) approaches \( 0 \) does not equal the function value at \( x = 0 \), there is a discontinuity at \( x = 0 \).
3. Type of Discontinuity:
- The limit exists and is finite, but it does not equal the function's value at \( x = 0 \). This indicates that there is a removable discontinuity at \( x = 0 \).
Therefore, we summarize:
- The function \( f \) is continuous on \( (-\infty, 0) \cup (0, \infty) \).
- There is a removable discontinuity at \( x = 0 \).
Hence, the correct answer is:
- The function \( f \) is continuous on \( (-\infty, 0) \cup (0, \infty) \).
- There is a removable discontinuity at [tex]\( x = 0 \)[/tex].
