Question

the function (f(x)=-2x^{3}+15x^{2}+84x - 3) is increasing on the interval ( , ), is decreasing on the interval ((-infty, )) and the interval ( , (infty)). the function has a relative maximum at ( , ).

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=- 2x^{3}+15x^{2}+84x - 3\). Using the power - rule \((x^n)^\prime=nx^{n - 1}\), we get \(f^\prime(x)=-6x^{2}+30x + 84\).

Step2: Set the derivative equal to zero

Set \(f^\prime(x)=0\), so \(-6x^{2}+30x + 84 = 0\). Divide through by \(-6\) to simplify: \(x^{2}-5x - 14=0\).

Step3: Factor the quadratic equation

Factor \(x^{2}-5x - 14\) as \((x - 7)(x+2)=0\). Then, by the zero - product property, \(x = 7\) or \(x=-2\).

Step4: Determine the intervals of increase and decrease

We consider the intervals \((-\infty,-2)\), \((-2,7)\) and \((7,\infty)\).
Test a value in the interval \((-\infty,-2)\), say \(x=-3\). Then \(f^\prime(-3)=-6\times(-3)^{2}+30\times(-3)+84=-54-90 + 84=-60<0\), so the function is decreasing on \((-\infty,-2)\).
Test a value in the interval \((-2,7)\), say \(x = 0\). Then \(f^\prime(0)=-6\times0^{2}+30\times0 + 84=84>0\), so the function is increasing on \((-2,7)\).
Test a value in the interval \((7,\infty)\), say \(x = 8\). Then \(f^\prime(8)=-6\times8^{2}+30\times8 + 84=-384+240 + 84=-60<0\), so the function is decreasing on \((7,\infty)\).
Since the function changes from increasing to decreasing at \(x = 7\), the relative maximum occurs at \(x = 7\).

Answer:

The function is increasing on the interval \((-2,7)\), decreasing on the intervals \((-\infty,-2)\) and \((7,\infty)\), and has a relative maximum at \(x = 7\).