Find values of \( a \) and \( b \) that make the function provided below both continuous and differentiable at 9 . \[ f(x)=\left\{\begin{array}{ll}a x+b & x<9 \\ x^{2} & x \geq 9\end{array}\right. \] \[ \begin{array}{l}a=\text { Number } \\ b=\text { Number }\end{array} \]

To ensure the function \( f(x) \) is continuous and differentiable at \( x = 9 \), we start by ensuring continuity. This requires that the left-hand limit \( \lim_{x \to 9^-} f(x) \) is equal to the right-hand limit \( \lim_{x \to 9^+} f(x) \). 1. **Continuity Condition:** \[ \lim_{x \to 9^-} f(x) = a(9) + b \] \[ \lim_{x \to 9^+} f(x) = 9^2 = 81 \] Setting them equal for continuity: \[ 9a + b = 81 \quad \text{(1)} \] 2. **Differentiability Condition:** For differentiability, the left-hand derivative must equal the right-hand derivative at \( x = 9 \). \[ f'(x) = a \quad \text{for } x < 9 \] \[ f'(x) = 2x \quad \text{for } x \geq 9 \] Evaluating at \( x = 9 \): \[ \lim_{x \to 9^-} f'(x) = a \] \[ \lim_{x \to 9^+} f'(x) = 2(9) = 18 \] Setting them equal for differentiability: \[ a = 18 \quad \text{(2)} \] 3. **Substituting \( a \) back into equation (1):** Substitute \( a = 18 \) into equation (1): \[ 9(18) + b = 81 \] \[ 162 + b = 81 \] \[ b = 81 - 162 = -81 \] Thus, the values are: \[ a = 18 \quad \text{and} \quad b = -81 \]