Question

the function (f(x)= - 4x^{3}-18x^{2}+120x - 8) is increasing on the interval ( , ). it is decreasing on the interval ((-infty,\text{ })) and the interval ((\text{ },infty)). the function has a relative maximum at (\text{ }.) attempt 1: 10 attempts remaining. submit answer next item

Explanation:

Step1: Find the derivative of the function

Given \(f(x)=- 4x^{3}-18x^{2}+120x - 8\), using the power - rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=-12x^{2}-36x + 120=-12(x^{2}+3x - 10)\).

Step2: Factor the derivative

Factor \(x^{2}+3x - 10=(x + 5)(x - 2)\), so \(f^\prime(x)=-12(x + 5)(x - 2)\).

Step3: Find the critical points

Set \(f^\prime(x)=0\), then \(-12(x + 5)(x - 2)=0\). Solving for \(x\), we get \(x=-5\) and \(x = 2\).

Step4: Determine the sign of the derivative on intervals

We consider the intervals \((-\infty,-5)\), \((-5,2)\) and \((2,\infty)\).

• For \(x\in(-\infty,-5)\), let \(x=-6\), then \(f^\prime(-6)=-12(-6 + 5)(-6 - 2)=-12\times(-1)\times(-8)=-96<0\), so the function is decreasing on \((-\infty,-5)\).
• For \(x\in(-5,2)\), let \(x = 0\), then \(f^\prime(0)=-12(0 + 5)(0 - 2)=-12\times5\times(-2)=120>0\), so the function is increasing on \((-5,2)\).
• For \(x\in(2,\infty)\), let \(x = 3\), then \(f^\prime(3)=-12(3 + 5)(3 - 2)=-12\times8\times1=-96<0\), so the function is decreasing on \((2,\infty)\).
• Since the function changes from increasing to decreasing at \(x = 2\), it has a relative maximum at \(x = 2\).

Answer:

The function has a relative maximum at \(x = 2\). It is decreasing on the interval \((-\infty,-5)\) and the interval \((2,\infty)\). It is increasing on the interval \((-5,2)\).