Increasing/Decreasing Intervals - Calculus 1

1. Determine the interval on which the function $f(x) = \arctan(x) - \frac{1}{2}\ln(1+x^2)$ is increasing.
   Answer: $(-\infty, 1)$ To find the interval of increase, we compute the derivative $f'(x)$ and find where it is positive. $f'(x) = \frac{d}{dx}(\arctan(x)) - \frac{1}{2}\frac{d}{dx}(\ln(1+x^2)) = \frac{1}{1+x^2} - \frac{1}{2} \cdot \frac{2x}{1+x^2} = \frac{1-x}{1+x^2}$. The function is increasing when $f'(x) > 0$. Since the denominator $1+x^2$ is always positive, the sign of $f'(x)$ is determined by the numerator $1-x$. The inequality $1-x > 0$ implies $1 > x$, or $x < 1$. Therefore, the function is increasing on the interval $(-\infty, 1)$.
2. Let $f(x)$ be a piecewise function defined as $f(x) = \begin{cases} x^2 + 2x + 2 & \text{if } x \le 0 \\ x^3 - 4x + 2 & \text{if } x > 0 \end{cases}$. On which of the following disjoint intervals is the function $f(x)$ decreasing?
   Answer: $(-\infty, -1)$ and $(0, \frac{2}{\sqrt{3}})$ We analyze the derivative $f'(x)$ on each piece. For $x < 0$, $f'(x) = 2x + 2$. This is negative when $2x+2 < 0$, i.e., $x < -1$. So $f$ is decreasing on $(-\infty, -1)$. For $x > 0$, $f'(x) = 3x^2 - 4$. This is negative when $3x^2 - 4 < 0$, i.e., $x^2 < 4/3$, which for $x>0$ means $0 < x < 2/\sqrt{3}$. So $f$ is decreasing on $(0, 2/\sqrt{3})$. The function is continuous at $x=0$. Combining the results, $f(x)$ is decreasing on $(-\infty, -1)$ and $(0, 2/\sqrt{3})$.
3. Let $f(x) = \frac{\ln(x)}{x^2}$. On which of the following intervals is $f(x)$ increasing?
   Answer: $(0, \sqrt{e})$ First, note the domain of $f(x)$ is $x > 0$. We use the quotient rule to find the derivative: $f'(x) = \frac{\frac{1}{x} \cdot x^2 - \ln(x) \cdot 2x}{(x^2)^2} = \frac{x - 2x\ln(x)}{x^4} = \frac{1 - 2\ln(x)}{x^3}$. The function is increasing where $f'(x) > 0$. Since $x > 0$, $x^3$ is positive, so the sign of $f'(x)$ is determined by the sign of the numerator, $1 - 2\ln(x)$. We need $1 - 2\ln(x) > 0$, which simplifies to $1 > 2\ln(x)$, or $\frac{1}{2} > \ln(x)$. Exponentiating both sides gives $e^{1/2} > x$, so $x < \sqrt{e}$. Combining with the domain $x>0$, the interval of increase is $(0, \sqrt{e})$.
4. On which of the following disjoint intervals is the function $f(x) = \frac{x^2+3}{x-1}$ increasing?
   Answer: $(-\infty, -1)$ and $(3, \infty)$ To find where $f(x)$ is increasing, we must find where $f'(x) > 0$. Using the quotient rule, $f'(x) = \frac{2x(x-1) - (x^2+3)(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2 - 3}{(x-1)^2} = \frac{x^2 - 2x - 3}{(x-1)^2} = \frac{(x-3)(x+1)}{(x-1)^2}$. The denominator $(x-1)^2$ is always positive for $x \neq 1$. Therefore, the sign of $f'(x)$ is determined by the numerator, $(x-3)(x+1)$. This is a parabola opening upwards with roots at $x=-1$ and $x=3$. The numerator is positive when $x < -1$ or $x > 3$. Thus, $f(x)$ is increasing on the intervals $(-\infty, -1)$ and $(3, \infty)$.
5. The function $f(x) = (x^2-1)|x-2|$ is decreasing on which of the following intervals?
   Answer: $(\frac{2 + \sqrt{7}}{3}, 2)$ We split the function into two cases. For $x < 2$, $f(x) = (x^2-1)(-(x-2)) = -x^3 + 2x^2 + x - 2$. Then $f'(x) = -3x^2 + 4x + 1$. The roots of $f'(x)=0$ are $x = \frac{-4 \pm \sqrt{16-4(-3)(1)}}{-6} = \frac{2 \pm \sqrt{7}}{3}$. Since $f'(x)$ is a downward-opening parabola, it is negative (and $f$ is decreasing) on $(-\infty, \frac{2 - \sqrt{7}}{3})$ and $(\frac{2 + \sqrt{7}}{3}, 2)$. For $x > 2$, $f(x) = (x^2-1)(x-2) = x^3 - 2x^2 - x + 2$. Then $f'(x) = 3x^2 - 4x - 1$. For $x>2$, $f'(x)$ is always positive, so $f$ is increasing on $(2, \infty)$. Thus, one of the intervals where $f$ is decreasing is $(\frac{2 + \sqrt{7}}{3}, 2)$.
6. For which range of values for the constant $k$ is the function $f(x) = x^3 - kx^2 + 3x + 1$ always increasing for all real numbers $x$?
   Answer: $-3 \le k \le 3$ For $f(x)$ to be always increasing, its derivative $f'(x)$ must be greater than or equal to zero for all $x$. The derivative is $f'(x) = 3x^2 - 2kx + 3$. This is a quadratic function that opens upwards. For it to be non-negative for all $x$, it can have at most one real root. This condition is met when the discriminant is less than or equal to zero. The discriminant is $\Delta = b^2 - 4ac = (-2k)^2 - 4(3)(3) = 4k^2 - 36$. Setting $\Delta \le 0$, we get $4k^2 - 36 \le 0$, which simplifies to $k^2 \le 9$. The solution to this inequality is $-3 \le k \le 3$.
7. Find the interval on which the function $f(x) = (x^2 - 4x)^{1/3}$ is increasing.
   Answer: $(2, \infty)$ First, we find the derivative using the chain rule: $f'(x) = \frac{1}{3}(x^2 - 4x)^{-2/3}(2x - 4) = \frac{2(x-2)}{3(x(x-4))^{2/3}}$. Critical points occur when the derivative is zero or undefined. $f'(x)=0$ when $x=2$. $f'(x)$ is undefined when $x=0$ and $x=4$. To find where $f(x)$ is increasing, we find where $f'(x) > 0$. The denominator involves a square, $(...)^{2/3}$, so it is always non-negative. Thus, the sign of $f'(x)$ is determined by the sign of the numerator, $2(x-2)$. The expression $2(x-2)$ is positive when $x > 2$. Therefore, the function is increasing on the interval $(2, \infty)$. Although the derivative is undefined at $x=4$, the function is continuous there, and since it is increasing on $(2, 4)$ and $(4, \infty)$, we can conclude it is increasing on $(2, \infty)$.
8. Let $f(x)$ be a twice-differentiable function such that $f'(x) > 0$ for all real numbers $x$. Let $g(x) = f(1-x^2)$. On which interval is $g(x)$ increasing?
   Answer: $(-\infty, 0)$ To determine where $g(x)$ is increasing, we must find where its derivative $g'(x)$ is positive. Using the chain rule, $g'(x) = f'(1-x^2) \cdot \frac{d}{dx}(1-x^2) = f'(1-x^2) \cdot (-2x)$. We are given that $f'(y) > 0$ for any input $y$. Therefore, the term $f'(1-x^2)$ is always positive. The sign of $g'(x)$ is thus determined by the sign of the other factor, $-2x$. The function $g(x)$ is increasing when $g'(x) > 0$, which means $-2x > 0$. Dividing by -2 and reversing the inequality sign gives $x < 0$. So, $g(x)$ is increasing on the interval $(-\infty, 0)$.
9. On which interval is the function $h(x) = x^2 e^{-x}$ increasing?
   Answer: $(0, 2)$ When you're asked where a function is increasing, you need to find where its derivative is positive. A function increases when its slope is positive and decreases when its slope is negative. For $h(x) = x^2 e^{-x}$, you'll need the product rule since this is the product of $x^2$ and $e^{-x}$. Taking the derivative: $h'(x) = 2x \cdot e^{-x} + x^2 \cdot (-e^{-x}) = e^{-x}(2x - x^2) = xe^{-x}(2 - x)$ Since $e^{-x}$ is always positive, the sign of $h'(x)$ depends only on $x(2-x)$. This expression equals zero when $x = 0$ or $x = 2$, giving you critical points that divide the number line into three intervals. Testing the sign on each interval: when $x < 0$, you have (negative)(positive) = negative; when $0 < x < 2$, you have (positive)(positive) = positive; when $x > 2$, you have (positive)(negative) = negative. Therefore, $h'(x) > 0$ only on $(0, 2)$, making this where the function increases. This confirms answer A is correct. Answer B suggests the function increases on $(-\infty, 0)$ and $(2, \infty)$, but these are exactly where $h'(x) < 0$. Answer C claims increasing on $(-\infty, 1)$, but the function decreases for $x < 0$. Answer D suggests increasing on $(2, \infty)$, but the derivative is negative there. Remember: always find the derivative first, then determine where it's positive. The critical points help you identify the intervals to test.
10. Let $f(x) = e^x \cos(x)$. On which of the following intervals is the function $f$ decreasing?
    Answer: $(\frac{\pi}{4}, \frac{5\pi}{4})$ To find where the function is decreasing, we need to find where its derivative is negative. Using the product rule, $f'(x) = \frac{d}{dx}(e^x \cos(x)) = e^x\cos(x) - e^x\sin(x) = e^x(\cos(x) - \sin(x))$. Since $e^x$ is always positive, the sign of $f'(x)$ is determined by the sign of $\cos(x) - \sin(x)$. The function $f$ is decreasing when $f'(x) < 0$, which means $\cos(x) - \sin(x) < 0$, or $\cos(x) < \sin(x)$. This inequality holds for $x$ in the interval $(\frac{\pi}{4}, \frac{5\pi}{4})$.

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