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

Explanation:

Step1: Find the derivative of the function

The function is (f(x)=x^{3}-x^{2}-7x). Using the power rule ((x^{n})^\prime = nx^{n - 1}), the derivative (f^\prime(x)=3x^{2}-2x - 7).

Step2: Find the critical points

Set (f^\prime(x)=0), so (3x^{2}-2x - 7 = 0). Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ax^{2}+bx + c = 0) (here (a = 3), (b=-2), (c=-7)), we have (x=\frac{2\pm\sqrt{(-2)^{2}-4\times3\times(-7)}}{2\times3}=\frac{2\pm\sqrt{4 + 84}}{6}=\frac{2\pm\sqrt{88}}{6}=\frac{2\pm2\sqrt{22}}{6}=\frac{1\pm\sqrt{22}}{3}). (x_1=\frac{1+\sqrt{22}}{3}\approx1.897), (x_2=\frac{1 - \sqrt{22}}{3}\approx-1.230).

Step3: Determine the intervals of increase and decrease

We use test - points in the intervals ((-\infty,-1.230)), ((-1.230,1.897)) and ((1.897,\infty)). For the interval ((-\infty,-1.230)), let (x=-2), then (f^\prime(-2)=3\times(-2)^{2}-2\times(-2)-7=12 + 4-7 = 9>0), so the function is increasing on ((-\infty,-1.230)). For the interval ((-1.230,1.897)), let (x = 0), then (f^\prime(0)=3\times0^{2}-2\times0-7=-7<0), so the function is decreasing on ([-1.230,1.897]). For the interval ((1.897,\infty)), let (x = 2), then (f^\prime(2)=3\times2^{2}-2\times2-7=12-4 - 7 = 1>0), so the function is increasing on ([1.897,\infty)).

Answer:

F. ([-1.230,1.897])