Suppose $f(x)$ has the following properties: - $f(x)$ is continuous and positive for att $x \in[6,19]$ - $f(x)$ is decreasing for all $x \in[6,19]$ - $\int_{6}^{19} f(x) d x=102$, - $\int_{6}^{12} f(x) d x=60$, Use this information to find the least and greatest possible value of $f(12)$. \[ \leq f(12) \leq \]

Step 1 ：Given that \( f(x) \) is continuous and positive for all \( x \in [6,19] \)

Step 2 ：Given that \( f(x) \) is decreasing for all \( x \in [6,19] \)

Step 3 ：Given that \( \int_{6}^{19} f(x) dx = 102 \)

Step 4 ：Given that \( \int_{6}^{12} f(x) dx = 60 \)

Step 5 ：The integral from 12 to 19 is \( \int_{12}^{19} f(x) dx = 102 - 60 = 42 \)

Step 6 ：The least possible value of \( f(12) \) occurs if \( f(x) \) decreases rapidly after 12, making the area from 12 to 19 as small as possible

Step 7 ：The greatest possible value of \( f(12) \) occurs if \( f(x) \) decreases slowly after 12, making the area from 12 to 19 as large as possible

Step 8 ：The least average value of \( f(x) \) from 12 to 19 is \( \frac{42}{7} = 6.0 \)

Step 9 ：The greatest average value of \( f(x) \) from 12 to 19 is \( \frac{42}{7} = 6.0 \)

Step 10 ：The least possible value of \( f(12) \) is equal to the least average value, which is 6.0

Step 11 ：The greatest possible value of \( f(12) \) is equal to the integral from 6 to 12 divided by the number of units, which is \( \frac{60}{6} = 10.0 \)

Step 12 ：Final Answer: \( \boxed{6 \leq f(12) \leq 10} \)