How do you find the maximum value of #f(x) = sinx ( 1+ cosx) #?

The maximum value is calculated with the first and second derivatives.

The function is

The first derivative is

When

The solutions are

The second derivative is

Therefore,

graph{sinx+1/2sin(2x) [-2.08, 10.404, -2.845, 3.395]}

To find the maximum value of $f(x) = \sin(x)(1 + \cos(x))$, we first need to find the critical points by taking the derivative of the function and setting it equal to zero. Then, we can determine if these points correspond to a maximum or minimum by checking the second derivative.

1. Take the derivative of $f(x)$: $f'(x) = (\sin(x))(1 + \cos(x)) + (\cos(x))(-\sin(x))$

2. Set the derivative equal to zero and solve for $x$: $(\sin(x))(1 + \cos(x)) + (\cos(x))(-\sin(x)) = 0$

3. Simplify and solve for $x$: $\sin(x) + \sin(x)\cos(x) - \cos(x)\sin(x) = 0$ $\sin(x) + \sin(x)\cos(x) - \sin(x)\cos(x) = 0$ $\sin(x) = 0$

4. Solve for $x$: $x = 0, \pi$

5. Determine the second derivative $f''(x)$: $f''(x) = (\cos(x))(1 + \cos(x)) - (\sin(x))^2$

6. Evaluate the second derivative at critical points $x = 0, \pi$: At $x = 0$: $f''(0) = (1)(1 + 1) - (0)^2 = 2 > 0$ (Minimum) At $x = \pi$: $f''(\pi) = (-1)(1 + (-1)) - (0)^2 = -2 < 0$ (Maximum)

Since $f''(\pi) < 0$, $f(x)$ has a maximum at $x = \pi$.

7. Find the maximum value of $f(x)$ by substituting $x = \pi$ into the original function: $f(\pi) = \sin(\pi)(1 + \cos(\pi))$ $f(\pi) = 0(1 + (-1)) = 0$

So, the maximum value of $f(x)$ is $0$ at $x = \pi$.