Question

(4) 5. given the graph of y = f(x), sketch the graph of its derivative y = f(x).

Explanation:

Step1: Identify increasing - decreasing intervals

Where \(y = f(x)\) is increasing, \(f'(x)>0\); where it is decreasing, \(f'(x)<0\). The function \(y = f(x)\) is decreasing on \((-\infty,- 1)\) and \((1,2)\), so \(f'(x)<0\) on these intervals. It is increasing on \((-1,1)\) and \((2,\infty)\), so \(f'(x)>0\) on these intervals.

Step2: Locate critical points

Critical points of \(y = f(x)\) occur where the slope is \(0\) or undefined. The function \(y = f(x)\) has critical points at \(x=-1,1,2\). At these points, \(f'(x) = 0\).

Step3: Analyze slope changes

Near \(x=-1\), the slope changes from negative to positive, so \(f'(x)\) crosses the \(x\) - axis from negative to positive at \(x = - 1\). Near \(x = 1\), the slope changes from positive to negative, so \(f'(x)\) crosses the \(x\) - axis from positive to negative at \(x = 1\). Near \(x = 2\), the slope changes from negative to positive, so \(f'(x)\) crosses the \(x\) - axis from negative to positive at \(x = 2\).

To sketch \(y = f'(x)\), we draw a curve that is below the \(x\) - axis on \((-\infty,-1)\) and \((1,2)\), above the \(x\) - axis on \((-1,1)\) and \((2,\infty)\), and crosses the \(x\) - axis at \(x=-1,1,2\).

Answer:

Sketch a curve for \(y = f'(x)\) that is negative on \((-\infty,-1)\cup(1,2)\), positive on \((-1,1)\cup(2,\infty)\) and crosses the \(x\) - axis at \(x=-1,1,2\).