The problem describes a function $f(x) = 1 - x \ln x$. It asks to find the domain of the function, calculate $\lim_{x \to 0^+} f(x)$ and $\lim_{x \to +\infty} f(x)$, and find vertical asymptotes. It also asks to determine if the curve touches the right side at $x=0$. Then it asks to find the derivative $f'(x)$, show if there is a maximum at $x = \frac{1}{e}$ and provide the table of variations. After that, it asks to find the intersection point between (C) and (D): $y = 1 - x$. Finally, calculate $f(1)$ and $f(2)$, and draw the curve (C) with $\ln 2 = 0.7$ and $\frac{1}{e} = 0.4$.

1. Problem Description

The problem describes a function $f(x) = 1 - x \ln x$. It asks to find the domain of the function, calculate $\lim_{x \to 0^+} f(x)$ and $\lim_{x \to +\infty} f(x)$, and find vertical asymptotes. It also asks to determine if the curve touches the right side at $x=0$.

Then it asks to find the derivative $f'(x)$, show if there is a maximum at $x = \frac{1}{e}$ and provide the table of variations.

After that, it asks to find the intersection point between (C) and (D): $y = 1 - x$.

Finally, calculate $f(1)$ and $f(2)$, and draw the curve (C) with $\ln 2 = 0.7$ and $\frac{1}{e} = 0.4$.

2. Solution Steps

(a) Domain:

Since we have $\ln x$ in the function, $x > 0$. Therefore, the domain of the function is $(0, +\infty)$.

Limit as $x$ approaches $0^+$:

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1 - x \ln x)$.

We need to find $\lim_{x \to 0^+} x \ln x$. We can rewrite this as $\lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}}$. This is in the indeterminate form $\frac{-\infty}{\infty}$, so we can apply L'Hopital's rule:

$\lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} (-x) = 0$.

Therefore, $\lim_{x \to 0^+} f(x) = 1 - 0 = 1$.

Limit as $x$ approaches $+\infty$:

$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (1 - x \ln x)$.

As $x$ approaches $+\infty$, $\ln x$ also approaches $+\infty$. Therefore, $x \ln x$ approaches $+\infty$.

So, $\lim_{x \to +\infty} (1 - x \ln x) = -\infty$.

Vertical Asymptotes:

Since the domain is $(0, +\infty)$ and $\lim_{x \to 0^+} f(x) = 1$, there is no vertical asymptote at $x=0$. Thus, there are no vertical asymptotes.

Curve touches the right side at $x=0$:

Since $\lim_{x \to 0^+} f(x) = 1$, the graph does not touch the right side at $x=0$.

(b) Derivative $f'(x)$:

$f(x) = 1 - x \ln x$

$f'(x) = 0 - (\ln x + x \cdot \frac{1}{x}) = -(\ln x + 1) = - \ln x - 1$.

Finding the maximum at $x = \frac{1}{e}$:

Set $f'(x) = 0$:

$-\ln x - 1 = 0$

$\ln x = -1$

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

Now we need to check the sign of $f'(x)$ around $x = \frac{1}{e}$:

If $x < \frac{1}{e}$, then $\ln x < -1$, so $-\ln x > 1$, and $f'(x) = -\ln x - 1 > 0$.

If $x > \frac{1}{e}$, then $\ln x > -1$, so $-\ln x < 1$, and $f'(x) = -\ln x - 1 < 0$.

Therefore, $f(x)$ has a maximum at $x = \frac{1}{e}$.

Table of variations:

x | 0 1/e +inf

-----------------------------------------------------

f'(x) | + 0 -

-----------------------------------------------------

f(x) | 1 increase f(1/e)=1+1/e decrease -inf

$f(\frac{1}{e}) = 1 - \frac{1}{e} \ln \frac{1}{e} = 1 - \frac{1}{e} (-1) = 1 + \frac{1}{e}$.

(c) Intersection between (C) and (D):

$y = f(x) = 1 - x \ln x$

$y = 1 - x$

$1 - x \ln x = 1 - x$

$x \ln x = x$

$\ln x = 1$ (assuming $x \ne 0$)

$x = e$

$y = 1 - e$

So the intersection point is $(e, 1-e)$.

(d) Calculate $f(1)$ and $f(2)$:

$f(1) = 1 - 1 \cdot \ln 1 = 1 - 1 \cdot 0 = 1$

$f(2) = 1 - 2 \ln 2 = 1 - 2(0.7) = 1 - 1.4 = -0.4$

3. Final Answer

(a) Domain: $(0, +\infty)$, $\lim_{x \to 0^+} f(x) = 1$, $\lim_{x \to +\infty} f(x) = -\infty$, no vertical asymptotes, the curve does not touch the right side at $x=0$.

(b) $f'(x) = - \ln x - 1$, maximum at $x = \frac{1}{e}$, $f(\frac{1}{e}) = 1 + \frac{1}{e}$. Table of variations as shown above.

(c) Intersection point: $(e, 1-e)$.

(d) $f(1) = 1$, $f(2) = -0.4$.