To determine the number of turning points of the function \( g(x) = 3x^3(x-5)^2(5-x)^4 \), we first need to analyze its degree and the nature of its roots.
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Degree of the Polynomial:
- The term \( 3x^3 \) contributes a degree of 3.
- The term \( (x-5)^2 \) contributes a degree of 2.
- The term \( (5-x)^4 \) also contributes a degree of 4.
Therefore, the total degree of \( g(x) \) is: \[ 3 + 2 + 4 = 9 \]
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Maximum Number of Turning Points:
- A polynomial of degree \( n \) can have at most \( n - 1 \) turning points. Therefore, for a polynomial of degree 9, the maximum number of turning points is: \[ 9 - 1 = 8 \]
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Multiplicity of the Roots:
- The roots of \( g(x) \) occur at \( x = 0 \), \( x = 5 \), and \( x = 5\) (counts due to the factor \( (5-x)^4 \)).
- The root \( x = 0 \) has multiplicity 3 (from \( 3x^3 \)).
- The root \( x = 5 \) has multiplicity 2 (from \( (x-5)^2 \)).
- The root \( x = 5 \) also has another multiplicity 4 (from \( (5-x)^4 \)). However, since it's the same root, we add the multiplicities:
- \( 2 + 4 = 6 \) for \( x = 5 \).
Thus, combining these:
- For \( x = 0 \): multiplicity 3 (it will turn down in between).
- For \( x = 5 \): multiplicity 6 (it will not change direction).
The general rule is that even multiplicities do not count as turning points but odd ones do. So, we count:
- The root at \( x = 0 \) contributes 2 turning points (as it crosses).
- The root at \( x = 5 \) contributes 0 turning points.
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Calculating the Total Turning Points: Since there are 2 turning points from \( x = 0 \) and none from \( x = 5 \): Therefore, considering the shape of the graph we see how it will behave with these turning points:
Thus, in summary, Rogelio should include 2 turning points in his sketch of the graph of \( g(x) \).
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