The problem asks us to sketch the graphs of two functions: (1) $y = x\sqrt{4-x^2}$ (2) $y = e^{-x} + \frac{x}{e} + 1$

1. Problem Description

The problem asks us to sketch the graphs of two functions:

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

(2) $y = e^{-x} + \frac{x}{e} + 1$

2. Solution Steps

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

Domain: The domain of this function is determined by the square root, requiring $4 - x^2 \geq 0$, which implies $x^2 \leq 4$, so $-2 \leq x \leq 2$.

Symmetry: Check for symmetry. $y(-x) = (-x)\sqrt{4-(-x)^2} = -x\sqrt{4-x^2} = -y(x)$. The function is odd.

Intercepts:

x-intercept: $y = 0 \Rightarrow x\sqrt{4-x^2} = 0 \Rightarrow x=0$ or $4-x^2 = 0 \Rightarrow x = \pm 2$. So the intercepts are $(0,0)$, $(2,0)$, and $(-2,0)$.

y-intercept: $x=0 \Rightarrow y = 0\sqrt{4-0} = 0$.

First Derivative:

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

Critical points: $y' = 0 \Rightarrow 4 - 2x^2 = 0 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}$.

If $x = \sqrt{2}$, $y = \sqrt{2}\sqrt{4-2} = \sqrt{2}\sqrt{2} = 2$.

If $x = -\sqrt{2}$, $y = -\sqrt{2}\sqrt{4-2} = -\sqrt{2}\sqrt{2} = -2$.

The critical points are $(\sqrt{2}, 2)$ and $(-\sqrt{2}, -2)$.

Second Derivative:

$y' = \frac{4-2x^2}{\sqrt{4-x^2}} = (4-2x^2)(4-x^2)^{-1/2}$.

$y'' = (-4x)(4-x^2)^{-1/2} + (4-2x^2)(-\frac{1}{2})(4-x^2)^{-3/2}(-2x) = \frac{-4x}{\sqrt{4-x^2}} + \frac{x(4-2x^2)}{(4-x^2)^{3/2}} = \frac{-4x(4-x^2) + x(4-2x^2)}{(4-x^2)^{3/2}} = \frac{-16x + 4x^3 + 4x - 2x^3}{(4-x^2)^{3/2}} = \frac{2x^3 - 12x}{(4-x^2)^{3/2}} = \frac{2x(x^2-6)}{(4-x^2)^{3/2}}$.

Since $-2 \le x \le 2$, $x^2 \le 4$, therefore $x^2 - 6$ will always be negative.

So, the sign of $y''$ is determined by the sign of $x$.

$y'' > 0$ when $x < 0$.

$y'' < 0$ when $x > 0$.

$y'' = 0$ when $x=0$.

The function is concave up for $-2 < x < 0$ and concave down for $0 < x < 2$. There is an inflection point at $(0, 0)$.

Max at $(\sqrt{2}, 2)$ and min at $(-\sqrt{2}, -2)$.

(2) $y = e^{-x} + \frac{x}{e} + 1$

First derivative:

$y' = -e^{-x} + \frac{1}{e}$.

Set $y' = 0$ to find critical points:

$-e^{-x} + \frac{1}{e} = 0$

$e^{-x} = \frac{1}{e} = e^{-1}$

$x = 1$.

At $x=1$, $y = e^{-1} + \frac{1}{e} + 1 = \frac{1}{e} + \frac{1}{e} + 1 = \frac{2}{e} + 1 \approx 1.736$.

So the critical point is $(1, \frac{2}{e} + 1)$.

Second derivative:

$y'' = e^{-x}$. Since $e^{-x} > 0$ for all $x$, the function is always concave up. This means the critical point at $x=1$ is a minimum.

As $x \to \infty$, $e^{-x} \to 0$, so $y \approx \frac{x}{e} + 1$, which is a straight line with positive slope.

As $x \to -\infty$, $e^{-x} \to \infty$, so $y$ goes to infinity.

3. Final Answer

(1) The graph of $y = x\sqrt{4-x^2}$ is defined for $-2 \le x \le 2$. It is an odd function with intercepts at $(0,0), (2,0), (-2,0)$. There is a local maximum at $(\sqrt{2}, 2)$ and a local minimum at $(-\sqrt{2}, -2)$. The function is concave up for $-2 < x < 0$ and concave down for $0 < x < 2$. There is an inflection point at $(0, 0)$.

(2) The graph of $y = e^{-x} + \frac{x}{e} + 1$ has a minimum at $(1, \frac{2}{e} + 1)$. The function is always concave up. As $x \to \infty$, $y$ approaches the line $y = \frac{x}{e} + 1$. As $x \to -\infty$, $y$ approaches infinity.