Kirk Hall
03/04/2023 · High School
19. \( x=\frac{1}{3} \sqrt{y}(y-3), \quad 1 \leqslant y \leqslant 9 \)
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Tutor-Verified Answer
Step-by-step Solution
To analyze the equation \( x = \frac{1}{3} \sqrt{y}(y - 3) \) for \( 1 \leq y \leq 9 \), we can start by examining the behavior of the function as \( y \) varies within the given range.
1. **Finding the endpoints**:
- When \( y = 1 \):
\[
x = \frac{1}{3} \sqrt{1}(1 - 3) = \frac{1}{3}(1)(-2) = -\frac{2}{3}
\]
- When \( y = 9 \):
\[
x = \frac{1}{3} \sqrt{9}(9 - 3) = \frac{1}{3}(3)(6) = 6
\]
2. **Finding critical points**:
To find the critical points, we can differentiate \( x \) with respect to \( y \):
\[
x = \frac{1}{3} \sqrt{y}(y - 3)
\]
Using the product rule, we differentiate:
\[
\frac{dx}{dy} = \frac{1}{3} \left( \frac{1}{2\sqrt{y}}(y - 3) + \sqrt{y} \cdot 1 \right)
\]
Simplifying this:
\[
\frac{dx}{dy} = \frac{1}{3} \left( \frac{y - 3}{2\sqrt{y}} + \sqrt{y} \right) = \frac{1}{3} \left( \frac{y - 3 + 2y}{2\sqrt{y}} \right) = \frac{1}{3} \left( \frac{3y - 3}{2\sqrt{y}} \right) = \frac{y - 1}{2\sqrt{y}}
\]
Setting \( \frac{dx}{dy} = 0 \):
\[
y - 1 = 0 \implies y = 1
\]
This is already one of our endpoints.
3. **Behavior of \( x \)**:
- For \( y = 1 \), we found \( x = -\frac{2}{3} \).
- For \( y = 9 \), we found \( x = 6 \).
- The derivative \( \frac{dx}{dy} = \frac{y - 1}{2\sqrt{y}} \) is positive for \( y > 1 \) and negative for \( y < 1 \). Thus, \( x \) is increasing in the interval \( (1, 9] \).
4. **Conclusion**:
The function \( x = \frac{1}{3} \sqrt{y}(y - 3) \) is increasing from \( -\frac{2}{3} \) at \( y = 1 \) to \( 6 \) at \( y = 9 \). Therefore, the range of \( x \) as \( y \) varies from \( 1 \) to \( 9 \) is:
\[
-\frac{2}{3} \leq x \leq 6
\]
Thus, the final answer is:
\[
\text{Range of } x: \left[-\frac{2}{3}, 6\right]
\]
Quick Answer
The range of \( x \) is \(\left[-\frac{2}{3}, 6\right]\).
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