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.
Understand the Problem
The question asks to find the critical points of the function f(x) = sin²x - sinx - 1 within the interval [0, 2π]. It also requires identifying where the function is increasing and decreasing, as well as determining the local and absolute extrema of the function.
Answer
Critical points: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$. Increasing: $(0, \frac{\pi}{6})$, $(\frac{5\pi}{6}, \frac{3\pi}{2})$. Decreasing: $(\frac{\pi}{6}, \frac{5\pi}{6})$, $(\frac{3\pi}{2}, 2\pi)$. Local min: $-\frac{3}{2}$ (at $x = \frac{\pi}{6}$), Absolute max: $0$ (at $x = \frac{3\pi}{2}$).
Answer for screen readers
- Critical points: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$
- Increasing on: $(0, \frac{\pi}{6})$ and $(\frac{5\pi}{6}, \frac{3\pi}{2})$
- Decreasing on: $(\frac{\pi}{6}, \frac{5\pi}{6})$ and $(\frac{3\pi}{2}, 2\pi)$
- Local minimum: $f(\frac{pi}{6}) = -\frac{3}{2}$
- Absolute maximum: $f(\frac{3\pi}{2}) = 0$
Steps to Solve
- 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:
$$ f'(x) = 2\sin x \cdot \cos x - \cos x = \cos x (2\sin x - 1) $$
- 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:
- $\cos x = 0$
- $2\sin x - 1 = 0$
- 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:
$$ x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2} $$
- 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 $(\frac{3\pi}{2}, 2\pi)$
- Evaluate function at critical points for extrema
Now we find $f$ at the critical points:
- $f(0) = \sin^2(0) - \sin(0) - 1 = -1$
- $f(\frac{\pi}{6}) = \sin^2(\frac{\pi}{6}) - \sin(\frac{\pi}{6}) - 1 = 0 - \frac{1}{2} - 1 = -\frac{3}{2}$
- $f(\frac{\pi}{2}) = \sin^2(\frac{\pi}{2}) - \sin(\frac{\pi}{2}) - 1 = 1 - 1 - 1 = -1$
- $f(\frac{5\pi}{6}) = \sin^2(\frac{5\pi}{6}) - \sin(\frac{5\pi}{6}) - 1 = \frac{3}{4} - \frac{1}{2} - 1 = -\frac{5}{4}$
- $f(\frac{3\pi}{2}) = \sin^2(\frac{3\pi}{2}) - \sin(\frac{3\pi}{2}) - 1 = 0 + 1 - 1 = 0$
- $f(2\pi) = \sin^2(2\pi) - \sin(2\pi) - 1 = 0 - 0 - 1 = -1$
- 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 $x = \frac{3\pi}{2}$.
- Critical points: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$
- Increasing on: $(0, \frac{\pi}{6})$ and $(\frac{5\pi}{6}, \frac{3\pi}{2})$
- Decreasing on: $(\frac{\pi}{6}, \frac{5\pi}{6})$ and $(\frac{3\pi}{2}, 2\pi)$
- Local minimum: $f(\frac{pi}{6}) = -\frac{3}{2}$
- Absolute maximum: $f(\frac{3\pi}{2}) = 0$
More Information
The function has its critical points derived from setting the first derivative to zero. The behavior of the function in terms of increasing and decreasing intervals helps identify local and absolute extremes. Such behavior is typical in sinusoidal functions due to their periodic nature.
Tips
- Forgetting to check endpoints when finding absolute extrema.
- Miscalculating the values of the function at critical points.
- Not testing intervals correctly might lead to incorrect increasing/decreasing intervals.