Answer
Critical number: x = -$\frac{2}{3}$
Increasing on (-$\infty$, -$\frac{2}{3}$)
Decreasing on (-$\frac{2}{3}$, $\infty$)
Relative maximum: (-$\frac{2}{3}$, -$\frac{2}{3}$)
Work Step by Step
f(x) = $-3x^{2}$ -4x-2
f'(x) = -6x-4
0 = -6x-4
Critical number: x = -$\frac{2}{3}$
Intervals:
(-$\infty$, -$\frac{2}{3}$) (-$\frac{2}{3}$, $\infty$)
Test Values: x = -2, x = 0
x= -2
f'(-2) = -6(-2)-4
f'(-2) = 8
Increasing on (-$\infty$, -$\frac{2}{3}$)
x = 0
f'(0) = -6(0) -4
f'(0) = -4
Decreasing on (-$\frac{2}{3}$, $\infty$)
f(-$\frac{2}{3}$) = -3 (-$\frac{2}{3})^{2}$ - 4 (-$\frac{2}{3}$) - 2
f(-$\frac{2}{3}$) =-$\frac{2}{3}$
Relative maximum: (-$\frac{2}{3}$, -$\frac{2}{3}$)
