Within the interval 0 ≤ x ≤ 2π, find the critical points of f(x) = sin²x - sinx - 1. Identify the open interval on which f is increasing and decreasing. Find the function's local a... Within the interval 0 ≤ x ≤ 2π, find the critical points of f(x) = sin²x - sinx - 1. Identify the open interval on which f is increasing and decreasing. Find the function's local and absolute extremes. Read more

Steps to Solve

1. Calculate the derivative of the function

To find critical points, we first need to find the derivative of the function $f(x) = \sin^2 x - \sin x - 1$. Using the chain rule and the derivative of $\sin x$, we have:

$$ f'(x) = 2\sin x \cdot \cos x - \cos x = \cos x (2\sin x - 1) $$

2. Set the derivative to zero

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

$$ \cos x (2\sin x - 1) = 0 $$

This gives us two conditions:

• $\cos x = 0$
• $2\sin x - 1 = 0$

3. Solve for critical points

For $\cos x = 0$, in the interval $[0, 2\pi]$, the solutions are:

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

For $2\sin x - 1 = 0$, we solve:

$$ \sin x = \frac{1}{2} $$

In the same interval, the solutions are:

$$ x = \frac{\pi}{6}, \frac{5\pi}{6} $$

Thus, the critical points are:

$$ x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2} $$

4. Determine increasing and decreasing intervals

We analyze the sign of the derivative $f'(x)$:

• Test intervals around the critical points:
  • Choose $x$ in $(0, \frac{\pi}{6})$: e.g., $x = 0$ => $f'(0) = 1 > 0 \Rightarrow$ increasing.
  • Choose $x$ in $(\frac{\pi}{6}, \frac{\pi}{2})$: e.g., $x = \frac{\pi}{4}$ => $f'(\frac{\pi}{4}) < 0 \Rightarrow$ decreasing.
  • Choose $x$ in $(\frac{\pi}{2}, \frac{5\pi}{6})$: e.g., $x = \frac{2\pi}{3}$ => $f'(\frac{2\pi}{3}) < 0 \Rightarrow$ decreasing.
  • Choose $x$ in $(\frac{5\pi}{6}, \frac{3\pi}{2})$: e.g., $x = \pi$ => $f'(\pi) = 1 > 0 \Rightarrow$ increasing.
  • Choose $x$ in $(\frac{3\pi}{2}, 2\pi)$: e.g., $x = \frac{7\pi}{6}$ => $f'(\frac{7\pi}{6}) < 0 \Rightarrow$ decreasing.

So the intervals are:

• Increasing: $(0, \frac{\pi}{6})$ and $(\frac{5\pi}{6}, \frac{3\pi}{2})$
• Decreasing: $(\frac{\pi}{6}, \frac{5\pi}{6})$ and $(\frac{3\pi}{2}, 2\pi)$

5. Evaluate function at critical points for extrema

Now we find $f$ at the critical points:

• $f(0) = \sin^2(0) - \sin(0) - 1 = -1$
• $f(\frac{\pi}{6}) = \sin^2(\frac{\pi}{6}) - \sin(\frac{\pi}{6}) - 1 = 0 - \frac{1}{2} - 1 = -\frac{3}{2}$
• $f(\frac{\pi}{2}) = \sin^2(\frac{\pi}{2}) - \sin(\frac{\pi}{2}) - 1 = 1 - 1 - 1 = -1$
• $f(\frac{5\pi}{6}) = \sin^2(\frac{5\pi}{6}) - \sin(\frac{5\pi}{6}) - 1 = \frac{3}{4} - \frac{1}{2} - 1 = -\frac{5}{4}$
• $f(\frac{3\pi}{2}) = \sin^2(\frac{3\pi}{2}) - \sin(\frac{3\pi}{2}) - 1 = 0 + 1 - 1 = 0$
• $f(2\pi) = \sin^2(2\pi) - \sin(2\pi) - 1 = 0 - 0 - 1 = -1$

6. Identify local and absolute extrema

From our evaluations:

• Local minimums: $f(\frac{\pi}{6}) = -\frac{3}{2}$ and $f(\frac{5\pi}{6}) = -\frac{5}{4}$
• Local maximum: $f(\frac{3\pi}{2}) = 0$

Considering the values calculated, the absolute minimum is $- \frac{3}{2}$ at $x = \frac{\pi}{6}$, and the absolute maximum is $0$ at $x = \frac{3\pi}{2}$.