Question

the function (f(x)=2x^{3}+12x^{2}-192x + 7) is decreasing on the interval ( ). it is increasing on the interval ((-infty,)) and the interval ((,infty)). the function has a local maximum at ( ).

Explanation:

Step1: Find the derivative of the function

Given \(f(x)=2x^{3}+12x^{2}-192x + 7\), using the power - rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=6x^{2}+24x-192 = 6(x^{2}+4x - 32)\).

Step2: Factor the derivative

Factor \(x^{2}+4x - 32=(x + 8)(x - 4)\), so \(f^\prime(x)=6(x + 8)(x - 4)\).

Step3: Find the critical points

Set \(f^\prime(x)=0\), then \(6(x + 8)(x - 4)=0\). The critical points are \(x=-8\) and \(x = 4\).

Step4: Determine the sign of the derivative in intervals

We consider the intervals \((-\infty,-8)\), \((-8,4)\) and \((4,\infty)\).

• For \(x\in(-\infty,-8)\), let \(x=-9\), then \(f^\prime(-9)=6(-9 + 8)(-9 - 4)=6\times(-1)\times(-13)=78>0\), so the function is increasing on \((-\infty,-8)\).
• For \(x\in(-8,4)\), let \(x = 0\), then \(f^\prime(0)=6(0 + 8)(0 - 4)=6\times8\times(-4)=-192<0\), so the function is decreasing on \((-8,4)\).
• For \(x\in(4,\infty)\), let \(x = 5\), then \(f^\prime(5)=6(5 + 8)(5 - 4)=6\times13\times1 = 78>0\), so the function is increasing on \((4,\infty)\).

Since the function changes from increasing to decreasing at \(x=-8\), it has a local maximum at \(x=-8\).

Answer:

The function is decreasing on the interval \((-8,4)\), increasing on the intervals \((-\infty,-8)\) and \((4,\infty)\), and has a local maximum at \(x=-8\).