Question

the function (f(x)=-2x^{3}-3x^{2}+180x - 8) is increasing on the interval ( , ). it is decreasing on the interval ((-infty,\text{ })) and the interval ( , (infty)). the function has a relative maximum at ( , ). next item

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 \(x = 5\).