Question

Rogelio is asked to sketch a graph of g(x)=3x^3(x−5)^2(5−x)^4 . How many turning points should his sketch include?

Answer 1

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.

1. **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:
\[
3 + 2 + 4 = 9
\]

2. **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:
\[
9 - 1 = 8
\]

3. **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.

4. **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 \( g(x) \).