The problem consists of several parts related to the function $f(x) = x^2 - x + a$, where 'a' is a constant. We need to calculate function values, find interpretations, analyze its derivative, determine its monotonicity, construct its table of variations, determine tangent lines and their equations, approximate function values, find the minimum value of the function, and demonstrate properties of tangents.

1. Problem Description

The problem consists of several parts related to the function $f(x) = x^2 - x + a$, where 'a' is a constant. We need to calculate function values, find interpretations, analyze its derivative, determine its monotonicity, construct its table of variations, determine tangent lines and their equations, approximate function values, find the minimum value of the function, and demonstrate properties of tangents.

2. Solution Steps

We will first calculate $f(0)$, $f(1)$, and $f(a)$ given the function $f(x) = x^2 - x + a$.

* Calculating $f(0)$:

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

* Calculating $f(1)$:

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

* Calculating $f(a)$:

$f(a) = (a)^2 - (a) + a = a^2$

The geometric interpretation of $f(0) = a$ and $f(1) = a$ is that the points $(0, a)$ and $(1, a)$ lie on the graph of the function $f(x)$. Since the y-values are the same, the line connecting those points is horizontal. The y-intercept is at $(0, a)$.

Next, we need to show that $f(x)$ is differentiable on $R$ and calculate $f'(x)$.

$f(x) = x^2 - x + a$.

Since $f(x)$ is a polynomial, it is differentiable on all real numbers R.

To find the derivative, we use the power rule:

$\frac{d}{dx}x^n = nx^{n-1}$.

$f'(x) = \frac{d}{dx}(x^2 - x + a) = 2x - 1 + 0 = 2x - 1$.

Now, we solve $f'(x) = 0$ for $x$:

$2x - 1 = 0$

$2x = 1$

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

Now we deduce the monotonicity of $f(x)$ on $R$:

Since $f'(x) = 2x - 1$, if $x < \frac{1}{2}$, then $f'(x) < 0$, which means $f(x)$ is decreasing.

If $x > \frac{1}{2}$, then $f'(x) > 0$, which means $f(x)$ is increasing.

Next, create the table of variations of $f(x)$.

$x$ ranges from $-\infty$ to $+\infty$.

$f'(x) = 0$ at $x = \frac{1}{2}$.

$f'(x)$ is negative for $x < \frac{1}{2}$ and positive for $x > \frac{1}{2}$.

Calculate $f(\frac{1}{2})$:

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

The function $f(x)$ has a tangent $(T)$ at $x=0$.

The equation of a tangent at a point $x_0$ is $y = f'(x_0)(x-x_0) + f(x_0)$.

At $x_0 = 0$, we have $f(0) = a$ and $f'(0) = 2(0) - 1 = -1$.

The equation of the tangent line $(T)$ at $x=0$ is:

$y = f'(0)(x - 0) + f(0) = -1(x - 0) + a = -x + a$.

Therefore, $y = -x + a$.

The affine tangent function $g$ to $f$ at $x=0$ is simply the tangent line equation we just found. So $g(x) = -x + a$.

To calculate an approximate value of $f(0.1)$, we can use the tangent line:

$f(0.1) \approx g(0.1) = -0.1 + a$.

We've already calculated $f'(x) = 2x - 1$.

The minimum value of $f(x)$ occurs at the vertex of the parabola, which is at $x = \frac{1}{2}$. The minimum value is $f(\frac{1}{2}) = a - \frac{1}{4}$.

To show that the tangent to $(C_f)$ at some point where its slope is zero is parallel to the x-axis (abscissa axis), consider $f'(x) = 2x-1$.

We want $f'(x) = 0$, which implies $x = \frac{1}{2}$. At $x = \frac{1}{2}$, the y-value is $f(\frac{1}{2}) = a - \frac{1}{4}$.

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

$y = f'(\frac{1}{2})(x - \frac{1}{2}) + f(\frac{1}{2}) = 0(x - \frac{1}{2}) + a - \frac{1}{4} = a - \frac{1}{4}$.

Thus, $y = a - \frac{1}{4}$. This is a horizontal line, which is parallel to the x-axis.

3. Final Answer

* $f(0) = a$

* $f(1) = a$

* $f(a) = a^2$

* $f'(x) = 2x - 1$

* $f'(x) = 0$ at $x = \frac{1}{2}$

* $f(x)$ is decreasing for $x < \frac{1}{2}$ and increasing for $x > \frac{1}{2}$

* Minimum value of $f(x)$ is $a - \frac{1}{4}$ at $x = \frac{1}{2}$

* Tangent equation at $x=0$: $y = -x + a$

* Approximate value of $f(0.1)$: $-0.1 + a$

* Tangent to $(C_f)$ where $f'(x)=0$ is $y = a - \frac{1}{4}$