2.9 Q-1

Question Statement

Determine the intervals in which the function $f(x)$ is increasing or decreasing within the given domains:

1. $f(x) = \sin x$, $x \in [-\pi, \pi]$
2. $f(x) = \cos x$, $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
3. $f(x) = 4 - x^2$, $x \in [-2, 2]$
4. $f(x) = x^2 + 3x + 2$, $x \in [-4, 1]$

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Background and Explanation

To analyze where a function is increasing or decreasing, we rely on its derivative $f'(x)$:

• If $f'(x) > 0$ in an interval, the function is increasing.
• If $f'(x) < 0$ in an interval, the function is decreasing.

We compute the derivative for each function and evaluate its sign over the specified intervals.

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Solution

1. $f(x) = \sin x$, $x \in [-\pi, \pi]$

• Derivative: $f'(x) = \cos x$

• Analysis:

  • $\cos x > 0$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so $f(x)$ is increasing here.
  • $\cos x < 0$ in $\left(-\pi, -\frac{\pi}{2}\right)$ and $\left(\frac{\pi}{2}, \pi\right)$, so $f(x)$ is decreasing in these intervals.

• Conclusion:

  • $f(x)$ is increasing in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
  • $f(x)$ is decreasing in $\left[-\pi, -\frac{\pi}{2}\right] \cup \left[\frac{\pi}{2}, \pi\right]$.

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2. $f(x) = \cos x$, $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

• Derivative: $f'(x) = -\sin x$

• Analysis:

  • $\sin x > 0$ in $(0, \frac{\pi}{2})$, so $f'(x) < 0$ and $f(x)$ is decreasing.
  • $\sin x < 0$ in $\left(-\frac{\pi}{2}, 0\right)$, so $f'(x) > 0$ and $f(x)$ is increasing.

• Conclusion:

  • $f(x)$ is increasing in $\left[-\frac{\pi}{2}, 0\right]$.
  • $f(x)$ is decreasing in $\left[0, \frac{\pi}{2}\right]$.

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3. $f(x) = 4 - x^2$, $x \in [-2, 2]$

• Derivative: $f'(x) = -2x$

• Analysis:

  • $f'(x) > 0$ when $x < 0$, so $f(x)$ is increasing in $(-2, 0)$.
  • $f'(x) < 0$ when $x > 0$, so $f(x)$ is decreasing in $(0, 2)$.

• Conclusion:

  • $f(x)$ is increasing in $[-2, 0]$.
  • $f(x)$ is decreasing in $[0, 2]$.

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4. $f(x) = x^2 + 3x + 2$, $x \in [-4, 1]$

• Derivative: $f'(x) = 2x + 3$

• Analysis:

  • $f'(x) < 0$ when $x < -\frac{3}{2}$, so $f(x)$ is decreasing in $(-4, -\frac{3}{2})$.
  • $f'(x) > 0$ when $x > -\frac{3}{2}$, so $f(x)$ is increasing in $(-\frac{3}{2}, 1)$.

• Conclusion:

  • $f(x)$ is decreasing in $[-4, -\frac{3}{2}]$.
  • $f(x)$ is increasing in $\left[-\frac{3}{2}, 1\right]$.

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Key Formulas or Methods Used

1. Derivative Test:
   • $f'(x) > 0$: Function is increasing.
   • $f'(x) < 0$: Function is decreasing.
2. Basic derivatives:
   • $\frac{d}{dx} (\sin x) = \cos x$
   • $\frac{d}{dx} (\cos x) = -\sin x$
   • $\frac{d}{dx} (4 - x^2) = -2x$
   • $\frac{d}{dx} (x^2 + 3x + 2) = 2x + 3$

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Summary of Steps

1. Compute the derivative $f'(x)$ for the given function.
2. Analyze the sign of $f'(x)$ within the specified intervals.
3. Identify where $f'(x) > 0$ (increasing) and $f'(x) < 0$ (decreasing).
4. Write conclusions for the intervals of increase and decrease.