Robbins Murphy
07/30/2023 · Primary School
b It is given that \( y=\frac{3 x^{2}-12}{5} \) For values of \( x \) in the range \( -3 \leq x \leq 2 \), find ii the smallest value of \( y \).
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Step-by-step Solution
To find the smallest value of \( y \) for the function \( y = \frac{3x^2 - 12}{5} \) in the range \( -3 \leq x \leq 2 \), we can follow these steps:
1. **Identify the function**:
\[
y = \frac{3x^2 - 12}{5}
\]
2. **Find the critical points**:
To find the critical points, we first need to take the derivative of \( y \) with respect to \( x \) and set it to zero.
\[
y' = \frac{d}{dx}\left(\frac{3x^2 - 12}{5}\right) = \frac{1}{5} \cdot (6x) = \frac{6x}{5}
\]
Setting the derivative equal to zero:
\[
\frac{6x}{5} = 0 \implies x = 0
\]
3. **Evaluate \( y \) at the critical point and endpoints**:
We need to evaluate \( y \) at the critical point \( x = 0 \) and at the endpoints \( x = -3 \) and \( x = 2 \).
- For \( x = 0 \):
\[
y(0) = \frac{3(0)^2 - 12}{5} = \frac{-12}{5} = -\frac{12}{5}
\]
- For \( x = -3 \):
\[
y(-3) = \frac{3(-3)^2 - 12}{5} = \frac{3(9) - 12}{5} = \frac{27 - 12}{5} = \frac{15}{5} = 3
\]
- For \( x = 2 \):
\[
y(2) = \frac{3(2)^2 - 12}{5} = \frac{3(4) - 12}{5} = \frac{12 - 12}{5} = \frac{0}{5} = 0
\]
4. **Compare the values**:
Now we compare the values of \( y \) at the critical point and the endpoints:
- \( y(0) = -\frac{12}{5} \)
- \( y(-3) = 3 \)
- \( y(2) = 0 \)
The smallest value of \( y \) is:
\[
-\frac{12}{5}
\]
Thus, the smallest value of \( y \) in the range \( -3 \leq x \leq 2 \) is \(\boxed{-\frac{12}{5}}\).
Quick Answer
The smallest value of \( y \) is \(-\frac{12}{5}\).
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