21. \( f(x)=-3 x^{2}+2 x-9 \) The function has one local max on its domain - because \( f^{\prime} \) changes sign from \( \frac{\text { Pos }}{} \) to hes as \( x \) increases through the critical point \( C \). Locate any local extrema.

To find the local extrema of the function \( f(x) = -3x^2 + 2x - 9 \), we first need to find its derivative: \[ f'(x) = -6x + 2. \] Setting the derivative equal to zero for critical points: \[ -6x + 2 = 0 \Rightarrow 6x = 2 \Rightarrow x = \frac{1}{3}. \] To determine whether this critical point is a maximum or a minimum, we can check the sign of the first derivative around \( x = \frac{1}{3} \): For \( x < \frac{1}{3} \) (pick \( x = 0 \)): \[ f'(0) = 2 > 0 \] (increasing) For \( x > \frac{1}{3} \) (pick \( x = 1 \)): \[ f'(1) = -4 < 0 \] (decreasing) Since \( f'(x) \) changes from positive to negative at \( x = \frac{1}{3} \), we can conclude that \( x = \frac{1}{3} \) is a local maximum. Now, plugging \( x = \frac{1}{3} \) back into the function to find the maximum value: \[ f\left(\frac{1}{3}\right) = -3\left(\frac{1}{3}\right)^2 + 2\left(\frac{1}{3}\right) - 9 = -3 \cdot \frac{1}{9} + \frac{2}{3} - 9 \] \[ = -\frac{1}{3} + \frac{2}{3} - 9 = \frac{1}{3} - 9 = -\frac{26}{3}. \] Thus, the local maximum occurs at \( \left(\frac{1}{3}, -\frac{26}{3}\right) \).