Find the two \( x \)-intercepts of the function \( f \) and show that \( f^{\prime}(x)=0 \) at some point between the two \( x \)-intercepts. \[ f(x)=-9 x \sqrt{x+9} \] smaller \( x \)-value \( \quad(x, y)=(\square) \) larger \( x \)-value \( \quad(x, y)=(\square) \) Find a value of \( x \) such that \( f^{\prime}(x)=0 \). \( x=\square \)

To find the \( x \)-intercepts of the function \( f(x) = -9x\sqrt{x + 9} \), we need to set \( f(x) = 0 \): \[ -9x\sqrt{x + 9} = 0 \] This occurs when either \( -9x = 0 \) or \( \sqrt{x + 9} = 0 \). 1. From \( -9x = 0 \), we get \( x = 0 \). 2. From \( \sqrt{x + 9} = 0 \), squaring both sides gives \( x + 9 = 0 \), or \( x = -9 \). Thus, the \( x \)-intercepts are: Smaller \( x \)-value \( \quad(x, y) = (-9, 0) \) Larger \( x \)-value \( \quad(x, y) = (0, 0) \) Next, we find \( f^{\prime}(x) \) and set it to zero. To differentiate, we can use the product and chain rules: First, express \( f(x) \) in terms of exponent: \[ f(x) = -9x(x + 9)^{1/2} \] Using the product rule: \[ f^{\prime}(x) = -9\left[ x\cdot\frac{1}{2}(x + 9)^{-1/2}\cdot1 + (x + 9)^{1/2}\cdot1 \right] \] Simplifying this results in: \[ f^{\prime}(x) = -9\left( \frac{x}{2\sqrt{x + 9}} + \sqrt{x + 9} \right) \] Setting \( f^{\prime}(x) = 0 \): \[ \frac{x}{2\sqrt{x + 9}} + \sqrt{x + 9} = 0 \] Solving for \( x \): \[ \frac{x + 2(x + 9)}{2\sqrt{x + 9}} = 0 \] \[ x + 2x + 18 = 3x + 18 = 0 \implies 3x = -18 \implies x = -6 \] Thus, a value of \( x \) such that \( f^{\prime}(x) = 0 \) is: \( x = -6 \) So, the final answers are: Smaller \( x \)-value \( \quad(x, y) = (-9, 0) \) Larger \( x \)-value \( \quad(x, y) = (0, 0) \) \( x = -6 \)