The problem asks us to determine the concavity of the given curves and find their inflection points, if any exist. (1) $y = xe^{-x}$ (2) $y = x - \cos x$ for $0 < x < 2\pi$ (3) $y = -x^4 + 4x^3 - 6x^2 + 4x$

1. Problem Description

The problem asks us to determine the concavity of the given curves and find their inflection points, if any exist.

(1) $y = xe^{-x}$

(2) $y = x - \cos x$ for $0 < x < 2\pi$

(3) $y = -x^4 + 4x^3 - 6x^2 + 4x$

2. Solution Steps

(1) $y = xe^{-x}$

First, find the first derivative $y'$:

$y' = e^{-x} + x(-e^{-x}) = e^{-x} - xe^{-x} = e^{-x}(1-x)$

Next, find the second derivative $y''$:

$y'' = -e^{-x}(1-x) + e^{-x}(-1) = -e^{-x} + xe^{-x} - e^{-x} = e^{-x}(x-2)$

To find inflection points, set $y'' = 0$:

$e^{-x}(x-2) = 0$

Since $e^{-x}$ is always positive, $x-2 = 0$, which gives $x = 2$.

Now, we analyze the sign of $y''$ around $x=2$:

- If $x < 2$, then $x-2 < 0$, so $y'' < 0$ (concave down).

- If $x > 2$, then $x-2 > 0$, so $y'' > 0$ (concave up).

The inflection point is $(2, 2e^{-2})$.

Concave down for $x<2$, concave up for $x>2$.

(2) $y = x - \cos x$ for $0 < x < 2\pi$

First, find the first derivative $y'$:

$y' = 1 + \sin x$

Next, find the second derivative $y''$:

$y'' = \cos x$

To find inflection points, set $y'' = 0$:

$\cos x = 0$

In the interval $0 < x < 2\pi$, the solutions are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.

Now, we analyze the sign of $y''$ around these points:

- If $0 < x < \frac{\pi}{2}$, then $\cos x > 0$, so $y'' > 0$ (concave up).

- If $\frac{\pi}{2} < x < \frac{3\pi}{2}$, then $\cos x < 0$, so $y'' < 0$ (concave down).

- If $\frac{3\pi}{2} < x < 2\pi$, then $\cos x > 0$, so $y'' > 0$ (concave up).

The inflection points are $(\frac{\pi}{2}, \frac{\pi}{2})$ and $(\frac{3\pi}{2}, \frac{3\pi}{2})$.

Concave up for $0 < x < \frac{\pi}{2}$, concave down for $\frac{\pi}{2} < x < \frac{3\pi}{2}$, and concave up for $\frac{3\pi}{2} < x < 2\pi$.

(3) $y = -x^4 + 4x^3 - 6x^2 + 4x$

First, find the first derivative $y'$:

$y' = -4x^3 + 12x^2 - 12x + 4$

Next, find the second derivative $y''$:

$y'' = -12x^2 + 24x - 12 = -12(x^2 - 2x + 1) = -12(x-1)^2$

To find inflection points, set $y'' = 0$:

$-12(x-1)^2 = 0$

This gives $x = 1$.

Now, we analyze the sign of $y''$ around $x=1$:

- If $x < 1$, then $(x-1)^2 > 0$, so $y'' < 0$ (concave down).

- If $x > 1$, then $(x-1)^2 > 0$, so $y'' < 0$ (concave down).

Since the concavity does not change at $x=1$, there is no inflection point. However, since $y''(1)=0$, we still consider $x=1$.

Concave down for all $x$. There is no inflection point.

3. Final Answer

(1) Concave down for $x < 2$, concave up for $x > 2$. Inflection point: $(2, 2e^{-2})$.

(2) Concave up for $0 < x < \frac{\pi}{2}$, concave down for $\frac{\pi}{2} < x < \frac{3\pi}{2}$, and concave up for $\frac{3\pi}{2} < x < 2\pi$. Inflection points: $(\frac{\pi}{2}, \frac{\pi}{2})$ and $(\frac{3\pi}{2}, \frac{3\pi}{2})$.

(3) Concave down for all $x$. No inflection point.