To analyze the statements regarding the graph of \( f(x) = \sec(x) \), let's evaluate each one:
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\( (0, 1) \) is a point on the graph.
- True. At \( x = 0 \), \( f(0) = \sec(0) = \frac{1}{\cos(0)} = 1 \). So, \( (0, 1) \) is indeed a point on the graph.
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\( f(x) \) is defined for all \( x \).
- False. The secant function is defined as \( \sec(x) = \frac{1}{\cos(x)} \). It is undefined whenever \( \cos(x) = 0 \), which occurs at odd multiples of \( \frac{\pi}{2} \).
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There is a vertical asymptote at \( x = \frac{\pi}{2} \).
- True. As \( x \) approaches \( \frac{\pi}{2} \) from the left, \( \cos(x) \) approaches 0, causing \( \sec(x) \) to approach infinity. Thus, there is a vertical asymptote at \( x = \frac{\pi}{2} \).
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\( f(x) \) is undefined when \( \sin(x) = 0 \).
- False. The secant function is not directly undefined when \( \sin(x) = 0 \). In fact, when \( \sin(x) = 0 \) (which occurs at integer multiples of \( \pi \)), \( \sec(x) \) is defined because \( \cos(x) \) will be either \( 1 \) or \( -1 \) (and both values are not equal to 0).
-
All \( x \)-values are included in the domain.
- False. Since \( f(x) = \sec(x) \) is undefined at \( \frac{\pi}{2} + n\pi \) for any integer \( n \), not all \( x \)-values are included in the domain.
Summary of True Statements:
- (1): \( (0, 1) \) is a point on the graph.
- (3): There is a vertical asymptote at \( x = \frac{\pi}{2} \).
So, the true statements from the options provided are statements 1 and 3.
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