We are given the function $f(x) = \ln(x^2 + x)$. We need to find the domain of $f$, the x-intercepts, the limit of $f(x)$ as $x$ approaches an accumulation point of the domain that is not in the domain, the limit of $f(x)$ as $x$ approaches $\pm \infty$, the first and second derivatives, the intervals of increase and decrease, the concavity, and sketch a graph. Then we have two limit problems. Then we have a problem using logarithmic differentiation and a final word problem.

1. Problem Description

We are given the function $f(x) = \ln(x^2 + x)$. We need to find the domain of $f$, the x-intercepts, the limit of $f(x)$ as $x$ approaches an accumulation point of the domain that is not in the domain, the limit of $f(x)$ as $x$ approaches $\pm \infty$, the first and second derivatives, the intervals of increase and decrease, the concavity, and sketch a graph. Then we have two limit problems. Then we have a problem using logarithmic differentiation and a final word problem.

2. Solution Steps

Question B2:

1. Domain of $f$: The domain of the natural logarithm function $\ln(u)$ is $u > 0$. Therefore, we need $x^2 + x > 0$, which means $x(x+1) > 0$. This inequality holds when $x < -1$ or $x > 0$. So the domain is $(-\infty, -1) \cup (0, \infty)$.

2. x-intercepts: We need to solve $f(x) = 0$, which means $\ln(x^2 + x) = 0$. Then $x^2 + x = e^0 = 1$, so $x^2 + x - 1 = 0$. Using the quadratic formula, $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$. Since $\frac{-1 - \sqrt{5}}{2} \approx \frac{-1 - 2.236}{2} \approx -1.618 < -1$ and $\frac{-1 + \sqrt{5}}{2} \approx \frac{-1 + 2.236}{2} \approx 0.618 > 0$, both roots are in the domain.

3. Limits and asymptotes: The boundary points of the domain are $x = -1$ and $x = 0$.

As $x \to -1^-$, $x^2 + x \to 0^+$, so $\lim_{x \to -1^-} \ln(x^2 + x) = -\infty$. Therefore, $x = -1$ is a vertical asymptote.

As $x \to 0^+$, $x^2 + x \to 0^+$, so $\lim_{x \to 0^+} \ln(x^2 + x) = -\infty$. Therefore, $x = 0$ is a vertical asymptote.

4. Limits as $x \to \pm \infty$: As $x \to \pm \infty$, $x^2 + x \to \infty$, so $\lim_{x \to \pm \infty} \ln(x^2 + x) = \infty$.

5. Derivatives:

a) $f'(x) = \frac{1}{x^2 + x} \cdot (2x + 1) = \frac{2x+1}{x^2 + x}$.

b) $f''(x) = \frac{(2)(x^2+x) - (2x+1)(2x+1)}{(x^2+x)^2} = \frac{2x^2 + 2x - (4x^2 + 4x + 1)}{(x^2+x)^2} = \frac{-2x^2 - 2x - 1}{(x^2+x)^2}$.

6. Intervals of increase and decrease:

$f'(x) = \frac{2x+1}{x(x+1)}$. We need to check where $f'(x) > 0$ and $f'(x) < 0$. The critical points are $x = -1, x = -1/2, x = 0$.

Since the domain is $(-\infty, -1) \cup (0, \infty)$, we check the intervals $(-\infty, -1)$ and $(0, \infty)$.

If $x < -1$, $2x+1 < 0$, $x < 0$, $x+1 < 0$, so $f'(x) = \frac{(-)}{(-)(-)} = \frac{(-)}{(+)} < 0$.

If $x > 0$, $2x+1 > 0$, $x > 0$, $x+1 > 0$, so $f'(x) = \frac{(+)}{(+)(+)} = \frac{(+)}{(+)} > 0$.

Therefore, $f(x)$ is decreasing on $(-\infty, -1)$ and increasing on $(0, \infty)$.

7. Concavity:

$f''(x) = \frac{-2x^2 - 2x - 1}{(x^2+x)^2}$. Since $(x^2+x)^2 > 0$, the sign of $f''(x)$ is determined by $-2x^2 - 2x - 1$. The discriminant is $(-2)^2 - 4(-2)(-1) = 4 - 8 = -4 < 0$. Since the leading coefficient is negative, $-2x^2 - 2x - 1 < 0$ for all $x$. Therefore, $f''(x) < 0$ for all $x$ in the domain, so $f(x)$ is concave down on $(-\infty, -1)$ and $(0, \infty)$.

8. Sketch: The graph has vertical asymptotes at $x = -1$ and $x = 0$. It is decreasing on $(-\infty, -1)$ and increasing on $(0, \infty)$. It is concave down everywhere. The x-intercepts are $x = \frac{-1 \pm \sqrt{5}}{2}$.

Question B3:

a) (i) $\lim_{x \to 0} \frac{\sin^{20} x}{x^{20} + \sin^{20} x} = \lim_{x \to 0} \frac{(\frac{\sin x}{x})^{20}}{1 + (\frac{\sin x}{x})^{20}} = \frac{1^{20}}{1 + 1^{20}} = \frac{1}{1+1} = \frac{1}{2}$.

(ii) $\lim_{x \to -\infty} \frac{\sqrt[4]{16x^4 + x^3 - 2x^2 + 1}}{x + 2025} = \lim_{x \to -\infty} \frac{\sqrt[4]{x^4(16 + \frac{1}{x} - \frac{2}{x^2} + \frac{1}{x^4})}}{x(1 + \frac{2025}{x})} = \lim_{x \to -\infty} \frac{|x| \sqrt[4]{16 + \frac{1}{x} - \frac{2}{x^2} + \frac{1}{x^4}}}{x(1 + \frac{2025}{x})}$. Since $x \to -\infty$, $|x| = -x$.

Therefore, $\lim_{x \to -\infty} \frac{-x \sqrt[4]{16 + \frac{1}{x} - \frac{2}{x^2} + \frac{1}{x^4}}}{x(1 + \frac{2025}{x})} = \lim_{x \to -\infty} \frac{-\sqrt[4]{16 + \frac{1}{x} - \frac{2}{x^2} + \frac{1}{x^4}}}{1 + \frac{2025}{x}} = \frac{-\sqrt[4]{16}}{1} = -2$.

b) $y = 2^{x^3} \sqrt{x^2 - 1} (x^4 - 2x^2)^3$.

Take the natural logarithm of both sides:

$\ln(y) = \ln(2^{x^3} \sqrt{x^2 - 1} (x^4 - 2x^2)^3) = x^3 \ln(2) + \frac{1}{2} \ln(x^2 - 1) + 3 \ln(x^4 - 2x^2)$.

Differentiate both sides with respect to $x$:

$\frac{y'}{y} = 3x^2 \ln(2) + \frac{1}{2} \cdot \frac{2x}{x^2 - 1} + 3 \cdot \frac{4x^3 - 4x}{x^4 - 2x^2} = 3x^2 \ln(2) + \frac{x}{x^2 - 1} + \frac{12x^3 - 12x}{x^4 - 2x^2} = 3x^2 \ln(2) + \frac{x}{x^2 - 1} + \frac{12x(x^2 - 1)}{x^2(x^2 - 2)} = 3x^2 \ln(2) + \frac{x}{x^2 - 1} + \frac{12(x^2 - 1)}{x(x^2 - 2)}$.

$y' = y [3x^2 \ln(2) + \frac{x}{x^2 - 1} + \frac{12(x^2 - 1)}{x(x^2 - 2)}] = 2^{x^3} \sqrt{x^2 - 1} (x^4 - 2x^2)^3 [3x^2 \ln(2) + \frac{x}{x^2 - 1} + \frac{12(x^2 - 1)}{x(x^2 - 2)}]$.

c) Let $x$ be the number. Then $x = \sqrt{x} + 2$. So $\sqrt{x} = x - 2$. Squaring both sides, $x = (x-2)^2 = x^2 - 4x + 4$. So $x^2 - 5x + 4 = 0$. Factoring, $(x-4)(x-1) = 0$. So $x = 4$ or $x = 1$. If $x = 4$, then $\sqrt{4} + 2 = 2 + 2 = 4$, so $x = 4$ works. If $x = 1$, then $\sqrt{1} + 2 = 1 + 2 = 3 \neq 1$, so $x = 1$ does not work.

Therefore, the number is

4.

3. Final Answer

Question B2:

1. Domain: $(-\infty, -1) \cup (0, \infty)$

2. x-intercepts: $x = \frac{-1 \pm \sqrt{5}}{2}$

3. Limit: $\lim_{x \to -1^-} f(x) = -\infty$, $\lim_{x \to 0^+} f(x) = -\infty$. Vertical asymptotes: $x = -1, x = 0$

4. Limit: $\lim_{x \to \pm \infty} f(x) = \infty$

5. a) $f'(x) = \frac{2x+1}{x^2+x}$

b) $f''(x) = \frac{-2x^2 - 2x - 1}{(x^2+x)^2}$

6. Decreasing on $(-\infty, -1)$ and increasing on $(0, \infty)$

7. Concave down on $(-\infty, -1)$ and $(0, \infty)$

8. Sketch: (see explanation above)

Question B3:

a) (i) $\frac{1}{2}$

(ii) $-2$

b) $y' = 2^{x^3} \sqrt{x^2 - 1} (x^4 - 2x^2)^3 [3x^2 \ln(2) + \frac{x}{x^2 - 1} + \frac{12(x^2 - 1)}{x(x^2 - 2)}]$

c) 4