Question

Consider the function on the interval \((0,2 \pi)\) For each function, (a) find the open interval(s) on which the function is increasing or decreasing, (b) apply the First Derivative Test to identify all relative extrema, and (c) use a graphing utility to confirm your results. $$ f(x)=x+2 \sin x $$

Short answer

The function \(f(x)=x+2 \sin x\) is increasing on the intervals \((0, \frac{2}{3} \pi)\) and \((\frac{4}{3} \pi , 2 \pi )\) and decreasing on the interval \((\frac{2}{3} \pi , \frac{4}{3} \pi )\). The function has a relative maxima at \(x= \frac{2}{3}\pi\) and a relative minima at \(x= \frac{4}{3}\pi\).

Step-by-step solution

Find the Derivative of the Function

The derivative of the function \(f(x)=x+2 \sin x\) is found by applying the standard rules of differentiation. The derivative of the function is \(f'(x)=1+2 \cos x\).

Find the Critical Points

Critical points occur where the derivative is equal to zero or undefined. Solve \(1+2 \cos x=0\). That gives us \(\cos x = -0.5\). Thus, the critical points on the interval \((0,2 \pi)\) are \(x= \frac{2}{3} \pi, \frac{4}{3} \pi\).

Identify Intervals of Increase or Decrease

Inspect the sign of the derivative on the intervals \( (0, \frac{2}{3} \pi ), ( \frac{2}{3} \pi , \frac{4}{3} \pi ), ( \frac{4}{3} \pi , 2 \pi ) \). If \(f'(x) > 0\), the function is increasing; if \(f'(x) < 0\), the function is decreasing. Thus, the function increases on \((0, \frac{2}{3} \pi)\) and \((\frac{4}{3} \pi , 2 \pi )\) and decreases on \((\frac{2}{3} \pi , \frac{4}{3} \pi )\).

Apply the First Derivative Test

By The First Derivative Test, \(f(x)\) has a relative maxima at \(x= \frac{2}{3}\pi\) and a relative minima at \(x= \frac{4}{3}\pi\).

Confirm Results with Graph

A graphical representation will confirm the findings. By plotting the function \(f(x)=x+2 \sin x\) on a graphing tool, you can confirm that the function has relative maxima and minima at the calculated points.