graph the function and identify intervals on which the function is increasing, decreasing, or…

graph the function and identify intervals on which the function is increasing, decreasing, or constant.\n\nf(x) = 3x³ - x² - 6x\n\nuse the graphing tool to graph the function.\n\nclick to enlarge graph

graph the function and identify intervals on which the function is increasing, decreasing, or constant.\n\nf(x) = 3x³ - x² - 6x\n\nuse the graphing tool to graph the function.\n\nclick to enlarge graph

Answer

Explanation:

Step1: Find the derivative of ( f(x) )

To determine where the function is increasing or decreasing, we first find the derivative of ( f(x) = 3x^3 - x^2 - 6x ). Using the power rule, the derivative ( f'(x) ) is: ( f'(x) = 9x^2 - 2x - 6 )

Step2: Find critical points

Set ( f'(x) = 0 ) to find critical points: ( 9x^2 - 2x - 6 = 0 ) Using the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) where ( a = 9 ), ( b = -2 ), and ( c = -6 ): ( x = \frac{2 \pm \sqrt{(-2)^2 - 4(9)(-6)}}{2(9)} = \frac{2 \pm \sqrt{4 + 216}}{18} = \frac{2 \pm \sqrt{220}}{18} = \frac{2 \pm 2\sqrt{55}}{18} = \frac{1 \pm \sqrt{55}}{9} ) Calculating the approximate values: ( \sqrt{55} \approx 7.416 ), so ( x_1 \approx \frac{1 + 7.416}{9} \approx 0.935 ) and ( x_2 \approx \frac{1 - 7.416}{9} \approx -0.713 )

Step3: Test intervals

We test the intervals determined by the critical points ( (-\infty, -0.713) ), ( (-0.713, 0.935) ), and ( (0.935, \infty) ) by plugging in test points into ( f'(x) ).

  • For ( x < -0.713 ) (e.g., ( x = -1 )): ( f'(-1) = 9(-1)^2 - 2(-1) - 6 = 9 + 2 - 6 = 5 > 0 ), so the function is increasing on ( (-\infty, \frac{1 - \sqrt{55}}{9}) ) (or approximately ( (-\infty, -0.713) )).
  • For ( -0.713 < x < 0.935 ) (e.g., ( x = 0 )): ( f'(0) = 0 - 0 - 6 = -6 < 0 ), so the function is decreasing on ( (\frac{1 - \sqrt{55}}{9}, \frac{1 + \sqrt{55}}{9}) ) (or approximately ( (-0.713, 0.935) )).
  • For ( x > 0.935 ) (e.g., ( x = 1 )): ( f'(1) = 9(1)^2 - 2(1) - 6 = 9 - 2 - 6 = 1 > 0 ), so the function is increasing on ( (\frac{1 + \sqrt{55}}{9}, \infty) ) (or approximately ( (0.935, \infty) )).

Answer:

The function ( f(x) = 3x^3 - x^2 - 6x ) is increasing on ( \left( -\infty, \frac{1 - \sqrt{55}}{9} \right) \cup \left( \frac{1 + \sqrt{55}}{9}, \infty \right) ) (or approximately ( (-\infty, -0.713) \cup (0.935, \infty) )) and decreasing on ( \left( \frac{1 - \sqrt{55}}{9}, \frac{1 + \sqrt{55}}{9} \right) ) (or approximately ( (-0.713, 0.935) )). There are no intervals where the function is constant.