The problem concerns the analysis of the function $f(x) = x^2 - x + \dots$. We are asked to: 1. Calculate $f(1)$, $f(0)$ and $f(2)$.

1. Problem Description

The problem concerns the analysis of the function $f(x) = x^2 - x + \dots$. We are asked to:

1. Calculate $f(1)$, $f(0)$ and $f(2)$.

2. Provide a geometric interpretation of the results from the previous question.

3. (a) Verify that $f$ is differentiable on $R$ and calculate $f'(x)$ for all $x \in R$.

(b) Solve the equation $f'(x) = 0$ in $R$ and deduce the monotonicity of $f$ on $R$.

(c) Sketch the table of variations of $f$.

4. (a) Verify that the curve $(C_f)$ admits a tangent $(T_A)$ at $\dots$.

(b) Show that $(T_A)$ has the equation $y = \dots$.

(c) Determine the affine function tangent $g$ to the function $f$ at $\dots$.

(d) Calculate an approximate value of the number $f(\dots)$.

5. (a) Calculate $f'( \dots )$.

(b) Show that $\dots$ is the minimal value of $f$ on $R$ by specifying the point where it occurs.

(c) Show that the tangent $(T)$ to $(C_f)$ at $\dots$ is parallel to the x-axis, and determine its equation.

6. Draw $(C_f)$ in the coordinate system $(O, \vec{i}, \vec{j})$ specifying $(T_A)$ and $(T)$.

Due to the image quality, some values are missing, and only a part of the questions can be solved. I will solve the parts of the problem that are readable.

2. Solution Steps

1. Calculate $f(1)$, $f(0)$ and $f(2)$.

Given $f(x) = x^2 - x + \dots$, let's denote the missing constant by $c$. So $f(x) = x^2 - x + c$.

$f(1) = (1)^2 - (1) + c = 1 - 1 + c = c$

$f(0) = (0)^2 - (0) + c = 0 - 0 + c = c$

$f(2) = (2)^2 - (2) + c = 4 - 2 + c = 2 + c$

2. Geometric Interpretation

Since $f(1) = f(0) = c$, it means that the points $(1, c)$ and $(0, c)$ lie on the curve of the function $f(x)$. Also, $f(2) = 2 + c$, which implies the point $(2, 2+c)$ lies on the curve of $f(x)$.

3. (a) Verify that $f$ is differentiable on $R$ and calculate $f'(x)$ for all $x \in R$.

Since $f(x) = x^2 - x + c$ is a polynomial function, it is differentiable on $R$. The derivative is:

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

3. (b) Solve the equation $f'(x) = 0$ in $R$ and deduce the monotonicity of $f$ on $R$.

We need to solve $f'(x) = 0$, i.e., $2x - 1 = 0$.

$2x = 1$

$x = \frac{1}{2}$

Now, we study the sign of $f'(x)$.

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

If $x > \frac{1}{2}$, then $2x > 1$, so $2x - 1 > 0$, thus $f'(x) > 0$.

Therefore, $f(x)$ is decreasing on $(-\infty, \frac{1}{2}]$ and increasing on $[\frac{1}{2}, \infty)$.

3. (c) Sketch the table of variations of $f$.

| x | -inf | 1/2 | +inf |

| :----- | :---------- | :---------- | :---------- |

| f'(x) | - | 0 | + |

| f(x) | \ decreasing | f(1/2) | / increasing |

$f(\frac{1}{2}) = (\frac{1}{2})^2 - \frac{1}{2} + c = \frac{1}{4} - \frac{1}{2} + c = -\frac{1}{4} + c$

5. (b) Show that $\dots$ is the minimal value of $f$ on $R$ by specifying the point where it occurs.

From the table of variations, we can deduce that $f(\frac{1}{2}) = -\frac{1}{4} + c$ is the minimum value of $f(x)$ on $R$, which is achieved at $x = \frac{1}{2}$.

5. (c) Show that the tangent $(T)$ to $(C_f)$ at $\dots$ is parallel to the x-axis, and determine its equation.

The tangent to $(C_f)$ is parallel to the x-axis when the derivative is zero. We already found that $f'(\frac{1}{2}) = 0$. So the tangent at $x=\frac{1}{2}$ is parallel to the x-axis.

The equation of the tangent at $x = \frac{1}{2}$ is:

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

$y = 0(x - \frac{1}{2}) + (-\frac{1}{4} + c)$

$y = -\frac{1}{4} + c$

3. Final Answer

1. $f(1) = c$, $f(0) = c$, $f(2) = 2 + c$

2. $(1, c)$, $(0, c)$ and $(2, 2+c)$ lie on the curve of $f(x)$.

3. (a) $f'(x) = 2x - 1$

(b) $x = \frac{1}{2}$, decreasing on $(-\infty, \frac{1}{2}]$, increasing on $[\frac{1}{2}, \infty)$.

(c) See the table of variations above.

4. (a) Unable to solve.

(b) Unable to solve.

(c) Unable to solve.

(d) Unable to solve.

5. (a) Unable to solve.

(b) The minimum value is $-\frac{1}{4} + c$ at $x = \frac{1}{2}$.

(c) The tangent at $x = \frac{1}{2}$ is parallel to the x-axis, and its equation is $y = -\frac{1}{4} + c$.

6. Unable to solve due to missing values.