We are given the function $f(x) = \frac{x^3}{(x+1)^2}$. The problem asks us to find the domain, intercepts, first and second derivatives, critical points, intervals of increase and decrease, points of inflection, asymptotes, and then to sketch the curve.

1. Problem Description

We are given the function $f(x) = \frac{x^3}{(x+1)^2}$. The problem asks us to find the domain, intercepts, first and second derivatives, critical points, intervals of increase and decrease, points of inflection, asymptotes, and then to sketch the curve.

2. Solution Steps

(a) Domain:

The function is defined for all $x$ except where the denominator is zero. The denominator $(x+1)^2$ is zero when $x+1=0$, so $x=-1$. Therefore, the domain of $f$ is all real numbers except $x=-1$. We can write this as $x \in (-\infty, -1) \cup (-1, \infty)$.

(b) Intercepts:

To find the x-intercepts, we set $f(x) = 0$. $\frac{x^3}{(x+1)^2} = 0$ implies $x^3 = 0$, so $x=0$. Thus, the x-intercept is $(0,0)$.

To find the y-intercepts, we set $x=0$. $f(0) = \frac{0^3}{(0+1)^2} = \frac{0}{1} = 0$. Thus, the y-intercept is $(0,0)$.

(c) First and second derivatives:

We use the quotient rule to find the first derivative $f'(x)$. The quotient rule is $\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

Here, $u = x^3$ and $v = (x+1)^2$. So, $\frac{du}{dx} = 3x^2$ and $\frac{dv}{dx} = 2(x+1)$.

$f'(x) = \frac{(x+1)^2(3x^2) - x^3(2(x+1))}{(x+1)^4} = \frac{(x+1)(3x^2(x+1) - 2x^3)}{(x+1)^4} = \frac{3x^3+3x^2-2x^3}{(x+1)^3} = \frac{x^3+3x^2}{(x+1)^3} = \frac{x^2(x+3)}{(x+1)^3}$

Now, we find the second derivative $f''(x)$. Again, we use the quotient rule with $u = x^3 + 3x^2$ and $v = (x+1)^3$.

Then $\frac{du}{dx} = 3x^2 + 6x$ and $\frac{dv}{dx} = 3(x+1)^2$.

$f''(x) = \frac{(x+1)^3(3x^2+6x) - (x^3+3x^2)(3(x+1)^2)}{(x+1)^6} = \frac{(x+1)^2[(x+1)(3x^2+6x) - (x^3+3x^2)(3)]}{(x+1)^6} = \frac{3x^3 + 6x^2 + 3x^2 + 6x - 3x^3 - 9x^2}{(x+1)^4} = \frac{6x}{(x+1)^4}$

(d) Critical points:

Critical points occur when $f'(x) = 0$ or $f'(x)$ is undefined. $f'(x) = \frac{x^2(x+3)}{(x+1)^3} = 0$ when $x^2(x+3) = 0$, which gives $x=0$ and $x=-3$. $f'(x)$ is undefined when $(x+1)^3 = 0$, which means $x=-1$. Since $x=-1$ is not in the domain of $f$, we only consider $x=0$ and $x=-3$. Thus, the critical points are $x=0$ and $x=-3$.

The critical points are $(-3, f(-3)) = (-3, \frac{(-3)^3}{(-3+1)^2}) = (-3, \frac{-27}{4})$ and $(0, f(0)) = (0,0)$.

(e) Intervals of increase and decrease:

We analyze the sign of $f'(x) = \frac{x^2(x+3)}{(x+1)^3}$.

- For $x < -3$, $x^2 > 0$, $x+3 < 0$, and $(x+1)^3 < 0$, so $f'(x) = \frac{(+)(-)}{(-)} = (+) > 0$.

- For $-3 < x < -1$, $x^2 > 0$, $x+3 > 0$, and $(x+1)^3 < 0$, so $f'(x) = \frac{(+)(+)}{(-)} = (-) < 0$.

- For $-1 < x < 0$, $x^2 > 0$, $x+3 > 0$, and $(x+1)^3 > 0$, so $f'(x) = \frac{(+)(+)}{(+)} = (+) > 0$.

- For $x > 0$, $x^2 > 0$, $x+3 > 0$, and $(x+1)^3 > 0$, so $f'(x) = \frac{(+)(+)}{(+)} = (+) > 0$.

Thus, $f$ is increasing on $(-\infty, -3)$ and $(-1, \infty)$, and decreasing on $(-3, -1)$.

(f) Points of inflection:

Points of inflection occur when $f''(x) = 0$ or $f''(x)$ is undefined and the concavity changes sign.

$f''(x) = \frac{6x}{(x+1)^4} = 0$ when $6x = 0$, so $x=0$. $f''(x)$ is undefined when $(x+1)^4=0$, or $x=-1$, which is not in the domain.

We check the sign of $f''(x)$ around $x=0$:

- For $x < 0$ (and $x > -1$), $f''(x) < 0$, so the graph is concave down.

- For $x > 0$, $f''(x) > 0$, so the graph is concave up.

Since the concavity changes at $x=0$, $(0,0)$ is a point of inflection.

(g) Asymptotes:

Vertical asymptotes occur where the function is undefined. Thus there is a vertical asymptote at $x=-1$.

To find horizontal asymptotes, we consider the limit as $x$ approaches infinity.

$\lim_{x \to \infty} \frac{x^3}{(x+1)^2} = \lim_{x \to \infty} \frac{x^3}{x^2+2x+1} = \lim_{x \to \infty} \frac{x}{1+2/x+1/x^2} = \infty$.

$\lim_{x \to -\infty} \frac{x^3}{(x+1)^2} = \lim_{x \to -\infty} \frac{x^3}{x^2+2x+1} = \lim_{x \to -\infty} \frac{x}{1+2/x+1/x^2} = -\infty$.

Since the limits are infinite, there are no horizontal asymptotes.

To check for slant asymptotes, we perform polynomial division:

$\frac{x^3}{x^2+2x+1} = x - 2 + \frac{3x+2}{x^2+2x+1}$. As $x \to \pm \infty$, $\frac{3x+2}{x^2+2x+1} \to 0$. Thus, there is a slant asymptote at $y=x-2$.

(h) Plot key points and sketch the curve.

Key points:

Intercept: (0,0)

Critical point: (-3, -27/4) = (-3, -6.75)

Inflection point: (0,0)

Asymptotes: x=-1, y=x-2

Increase: $(-\infty, -3), (-1, \infty)$

Decrease: $(-3, -1)$

3. Final Answer

(a) Domain: $x \in (-\infty, -1) \cup (-1, \infty)$

(b) Intercepts: (0,0)

(c) $f'(x) = \frac{x^2(x+3)}{(x+1)^3}$, $f''(x) = \frac{6x}{(x+1)^4}$

(d) Critical points: (-3, -27/4) and (0,0)

(e) Increasing: $(-\infty, -3) \cup (-1, \infty)$, Decreasing: $(-3, -1)$

(f) Inflection point: (0,0), Concave down for $x < 0, x \ne -1$, Concave up for $x > 0$

(g) Asymptotes: Vertical asymptote at $x = -1$, Slant asymptote at $y = x-2$

(h) The sketch is omitted, but should show the points, asymptotes and concavity.