1. **State the problem:** Find the range of the function $$h(x) = \frac{640}{x^2} + 0.5$$ for $$4 \leq x \leq 14$$. 2. **Understand the function:** The function is a sum of a rational term and a constant. Since $$x^2$$ is in the denominator, as $$x$$ increases, $$\frac{640}{x^2}$$ decreases. 3. **Find the values of $$h(x)$$ at the endpoints:** At $$x=4$$: $$h(4) = \frac{640}{4^2} + 0.5 = \frac{640}{16} + 0.5 = 40 + 0.5 = 40.5$$ At $$x=14$$: $$h(14) = \frac{640}{14^2} + 0.5 = \frac{640}{196} + 0.5 \approx 3.2653 + 0.5 = 3.7653$$ 4. **Determine the range:** Since $$h(x)$$ decreases as $$x$$ increases (because $$\frac{640}{x^2}$$ decreases), the maximum value is at $$x=4$$ and the minimum at $$x=14$$. Therefore, the range of $$h$$ is: $$[3.7653, 40.5]$$ --- 5. **Find $$h^{-1}(10)$$:** We want to find $$x$$ such that: $$h(x) = 10$$ Set up the equation: $$\frac{640}{x^2} + 0.5 = 10$$ Subtract 0.5 from both sides: $$\frac{640}{x^2} = 9.5$$ Multiply both sides by $$x^2$$: $$640 = 9.5 x^2$$ Divide both sides by 9.5: $$x^2 = \frac{640}{9.5}$$ Simplify: $$x^2 = \frac{640}{9.5} = \frac{640 \times 2}{19} = \frac{1280}{19} \approx 67.3684$$ Take the positive square root (since $$x \geq 4$$): $$x = \sqrt{67.3684} \approx 8.205$$ So, $$h^{-1}(10) \approx 8.205$$ 6. **Interpretation:** The value $$h^{-1}(10)$$ means the diameter $$x$$ of the cylinder when the height $$h(x)$$ is 10 cm. 7. **Range of $$h^{-1}$$:** Since $$h$$ maps $$[4,14]$$ to $$[3.7653,40.5]$$, the inverse function $$h^{-1}$$ maps $$[3.7653,40.5]$$ back to $$[4,14]$$. Therefore, the range of $$h^{-1}$$ is: $$[4,14]$$