To determine whether the graph of \( j(x) = \frac{12}{5} x^3 - \frac{27}{5} x \) is above or below the x-axis at the specified x-values (\( x = 0 \), \( x = \frac{3}{2} \), and \( x = -\frac{3}{2} \)), we can evaluate \( j(x) \) at each point and analyze the signs.
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At \( x = 0 \): \[ j(0) = \frac{12}{5}(0^3) - \frac{27}{5}(0) = 0 \]
- Result: \( j(0) = 0 \) → On the x-axis
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At \( x = \frac{3}{2} \): \[ j\left(\frac{3}{2}\right) = \frac{12}{5}\left(\frac{3}{2}\right)^3 - \frac{27}{5}\left(\frac{3}{2}\right) \] Calculate \( \left(\frac{3}{2}\right)^3 = \frac{27}{8} \): \[ j\left(\frac{3}{2}\right) = \frac{12}{5} \cdot \frac{27}{8} - \frac{27}{5} \cdot \frac{3}{2} \] \[ = \frac{12 \cdot 27}{40} - \frac{27 \cdot 3}{10} \] \[ = \frac{324}{40} - \frac{81}{10} = \frac{324}{40} - \frac{324}{40} = 0 \]
- Result: \( j\left(\frac{3}{2}\right) = 0 \) → On the x-axis
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At \( x = -\frac{3}{2} \): \[ j\left(-\frac{3}{2}\right) = \frac{12}{5}\left(-\frac{3}{2}\right)^3 - \frac{27}{5}\left(-\frac{3}{2}\right) \] Calculate \( \left(-\frac{3}{2}\right)^3 = -\frac{27}{8} \): \[ j\left(-\frac{3}{2}\right) = \frac{12}{5} \cdot \left(-\frac{27}{8}\right) + \frac{27}{5} \cdot \frac{3}{2} \] \[ = -\frac{324}{40} + \frac{81}{10} = -\frac{324}{40} + \frac{324}{40} = 0 \]
- Result: \( j\left(-\frac{3}{2}\right) = 0 \) → On the x-axis
Summary:
- For \( x < -\frac{3}{2} \): Above the x-axis
- For \( x = -\frac{3}{2} \): On the x-axis
- For \( -\frac{3}{2} < x < 0 \): Above the x-axis
- For \( x = 0 \): On the x-axis
- For \( 0 < x < \frac{3}{2} \): Above the x-axis
- For \( x = \frac{3}{2} \): On the x-axis
- For \( x > \frac{3}{2} \): Above the x-axis
So the graph will be on the x-axis at \( x = 0, x = \frac{3}{2}, \) and \( x = -\frac{3}{2} \), and will be above the x-axis in the other regions around these points.