the function (f(x)=-2x^{3}-3x^{2}+180x - 8) is increasing on the interval ( , ). it is decreasing on the…

Explanation:

Step1: Find the derivative of the function

Given (f(x)=- 2x^{3}-3x^{2}+180x - 8), using the power - rule ((x^n)^\prime=nx^{n - 1}), we get (f^\prime(x)=-6x^{2}-6x + 180).

Step2: Set the derivative equal to zero

(-6x^{2}-6x + 180 = 0). Divide through by (-6) to simplify: (x^{2}+x - 30=0).

Step3: Solve the quadratic equation

Factor the quadratic equation (x^{2}+x - 30=(x + 6)(x - 5)=0). So the critical points are (x=-6) and (x = 5).

Step4: Determine the sign of the derivative in intervals

We consider the intervals ((-\infty,-6)), ((-6,5)) and ((5,\infty)).

• For (x\in(-\infty,-6)), let's choose (x=-7). Then (f^\prime(-7)=-6\times(-7)^{2}-6\times(-7)+180=-6\times49 + 42+180=-294+42 + 180=-72<0), so the function is decreasing on ((-\infty,-6)).
• For (x\in(-6,5)), let's choose (x = 0). Then (f^\prime(0)=-6\times0^{2}-6\times0 + 180=180>0), so the function is increasing on ((-6,5)).
• For (x\in(5,\infty)), let's choose (x = 6). Then (f^\prime(6)=-6\times6^{2}-6\times6+180=-6\times36-36 + 180=-216-36 + 180=-72<0), so the function is decreasing on ((5,\infty)).
• Since the function changes from increasing to decreasing at (x = 5), the function has a relative maximum at (x = 5).

Answer:

The function is increasing on the interval ((-6,5)), decreasing on the intervals ((-\infty,-6)) and ((5,\infty)), and has a relative maximum at (x = 5).