To determine the number of discontinuities in the given piecewise function, we need to check the points at which the function changes its definition, specifically at \( x = -3 \), \( x = 3 \), and \( x = 5 \). We will assess the continuity of the function at those points.
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At \( x = -3 \):
- From the left (\( x < -3 \)): \( f(x) = 3e^x + 1 \)
- \( f(-3) = 3e^{-3} + 1 \) (a specific value).
- From the right (\( -3 < x \leq 3 \)): \( f(x) = \frac{2}{3}x^2 - 1 \)
- \( f(-3) = \frac{2}{3}(-3)^2 - 1 = \frac{2}{3} \cdot 9 - 1 = 6 - 1 = 5 \).
- Since \( 3e^{-3} + 1 \neq 5 \), there is a discontinuity at \( x = -3 \).
- From the left (\( x < -3 \)): \( f(x) = 3e^x + 1 \)
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At \( x = 3 \):
- From the left (\( -3 < x \leq 3 \)): \( f(x) = \frac{2}{3}x^2 - 1 \)
- \( f(3) = \frac{2}{3}(3)^2 - 1 = \frac{2}{3} \cdot 9 - 1 = 6 - 1 = 5 \).
- From the right (\( 3 < x < 5 \)): \( f(x) = -\frac{7}{2}x + \frac{29}{2} \)
- \( f(3) = -\frac{7}{2}(3) + \frac{29}{2} = -\frac{21}{2} + \frac{29}{2} = \frac{8}{2} = 4 \).
- Since \( 5 \neq 4 \), there is a discontinuity at \( x = 3 \).
- From the left (\( -3 < x \leq 3 \)): \( f(x) = \frac{2}{3}x^2 - 1 \)
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At \( x = 5 \):
- From the left (\( 3 < x < 5 \)): \( f(x) = -\frac{7}{2}x + \frac{29}{2} \)
- \( f(5) = -\frac{7}{2}(5) + \frac{29}{2} = -\frac{35}{2} + \frac{29}{2} = -\frac{6}{2} = -3 \).
- From the right (\( x \geq 5 \)): \( f(x) = \log(2x - 4) \)
- \( f(5) = \log(2 \cdot 5 - 4) = \log(10 - 4) = \log(6) \).
- Since \( -3 \neq \log(6) \), there is a discontinuity at \( x = 5 \).
- From the left (\( 3 < x < 5 \)): \( f(x) = -\frac{7}{2}x + \frac{29}{2} \)
Summary:
- The function has discontinuities at \( x = -3 \), \( x = 3 \), and \( x = 5 \).
Thus, the total number of discontinuities is 3. The correct response is:
three
1 answer