9 Given that the equation of a curve is \( y=\frac{10}{x}-x \), find the coordinates of the points on the curve at which the gradient is \( -\frac{7}{2} \).

To find the points on the curve where the gradient is \(-\frac{7}{2}\), we first need to determine the derivative of the function \(y = \frac{10}{x} - x\). Calculating the derivative: \[ y' = -\frac{10}{x^2} - 1 \] Now, we set this equal to \(-\frac{7}{2}\): \[ -\frac{10}{x^2} - 1 = -\frac{7}{2} \] Rearranging gives: \[ -\frac{10}{x^2} = -\frac{7}{2} + 1 \quad \Rightarrow \quad -\frac{10}{x^2} = -\frac{5}{2} \] Multiplying both sides by -1: \[ \frac{10}{x^2} = \frac{5}{2} \] Cross-multiplying gives: \[ 10 \cdot 2 = 5x^2 \quad \Rightarrow \quad 20 = 5x^2 \] Dividing by 5: \[ 4 = x^2 \quad \Rightarrow \quad x = 2 \text{ or } x = -2 \] Now, we substitute \(x = 2\) and \(x = -2\) back into the original equation to find the corresponding \(y\) values: For \(x = 2\): \[ y = \frac{10}{2} - 2 = 5 - 2 = 3 \quad \Rightarrow \quad (2, 3) \] For \(x = -2\): \[ y = \frac{10}{-2} - (-2) = -5 + 2 = -3 \quad \Rightarrow \quad (-2, -3) \] Thus, the coordinates of the points on the curve where the gradient is \(-\frac{7}{2}\) are \((2, 3)\) and \((-2, -3)\).