Mathematics - Application of Derivative Question with Solution | TestHub

Step 1: Find the derivative of $f(x)$

We start by differentiating the function:

$f^{\prime }(x)=\frac{d}{dx}(x+\sin (2x))=1+2\cos (2x)$

Step 2: Set the derivative equal to zero

To find the critical points, we set the derivative equal to zero:

$1+2\cos (2x)=0$

$2\cos (2x)=-1$

$\cos (2x)=-\frac{1}{2}$

Step 3: Solve for x

The solutions for $\cos (2x)=-\frac{1}{2}$ in the interval $(0,4\pi )$ (since $2x$ will range from $0$ to $4\pi$) are:

$2x=\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}$

Dividing by 2 gives us the critical points:

$x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$

Step 4: Evaluate the second derivative

Next, we find the second derivative to determine the nature of these critical points:

$f^{\prime \prime }(x)=\frac{d}{dx}(1+2\cos (2x))=-4\sin (2x)$

Step 5: Determine local maxima and minima

Now we evaluate $f^{\prime \prime }(x)$ at each critical point:

1. For $x=\frac{\pi }{3}$:

$f^{\prime \prime }(\frac{\pi }{3})=-4\sin (\frac{2\pi }{3})=-4\cdot \frac{\sqrt{3}}{2}=-2\sqrt{3}<0$ (local maximum)

2. For $x=\frac{2\pi }{3}$:

$f^{\prime \prime }(\frac{2\pi }{3})=-4\sin (\frac{4\pi }{3})=-4\cdot (-\frac{\sqrt{3}}{2})=2\sqrt{3}>0$ (local minimum)

3. For $x=\frac{4\pi }{3}$:

$f^{\prime \prime }(\frac{4\pi }{3})=-4\sin (\frac{8\pi }{3})=-4\cdot \frac{\sqrt{3}}{2}=-2\sqrt{3}<0$ (local maximum)

4. For $x=\frac{5\pi }{3}$:

$f^{\prime \prime }(\frac{5\pi }{3})=-4\sin (\frac{10\pi }{3})=-4\cdot (-\frac{\sqrt{3}}{2})=2\sqrt{3}>0$ (local minimum)

Step 6: Evaluate $f(x)$ at critical points and endpoints

Now we evaluate $f(x)$ at the critical points and the endpoints of the interval $(0,2\pi )$:

1. $f(0)=0+\sin (0)=0$

2. $f(2\pi )=2\pi +\sin (4\pi )=2\pi$

3. $f(\frac{\pi }{3})=\frac{\pi }{3}+\sin (\frac{2\pi }{3})=\frac{\pi }{3}+\frac{\sqrt{3}}{2}$

4. $f(\frac{2\pi }{3})=\frac{2\pi }{3}+\sin (\frac{4\pi }{3})=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$

5. $f(\frac{4\pi }{3})=\frac{4\pi }{3}+\sin (\frac{8\pi }{3})=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}$

6. $f(\frac{5\pi }{3})=\frac{5\pi }{3}+\sin (\frac{10\pi }{3})=\frac{5\pi }{3}-\frac{\sqrt{3}}{2}$

Step 7: Compare values

Now we compare all these values to find the maximum and minimum:

- $f(0)=0$

- $f(2\pi )=2\pi$

- $f(\frac{\pi }{3})=\frac{\pi }{3}+\frac{\sqrt{3}}{2}$

- $f(\frac{2\pi }{3})=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$

- $f(\frac{4\pi }{3})=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}$

- $f(\frac{5\pi }{3})=\frac{5\pi }{3}-\frac{\sqrt{3}}{2}$

Conclusion

The minimum value is 0 at $x=0$ and the maximum value is $2\pi$ at $x=2\pi$.