The provided image contains a mathematical problem related to polynomial functions and calculus. I will extract and analyze the content step by step to answer the subparts.

Aufgabe 1.1.2 Breakdown

The problem is centered around the function $f(x)$, a real-valued polynomial function with properties:

• Its graph is symmetrical about the $y$-axis, meaning $f(x)$ is an even function.
• Analysis of $f(x)$ and its derivatives $f'(x)$, $F(x)$ (antiderivative), and integration values is required.

Subtasks:

a)

"Begründen Sie, dass $f$ mindestens den Grad vier besitzt."

• $f(x)$ is symmetric about the $y$-axis, implying $f(x)$ is an even polynomial.
  • For even polynomials, the terms are of even powers (e.g., $x^4, x^2, x^0$).
• Looking at the graph, the behavior of $f(x)$ shows three turning points.
  • A polynomial's degree $n$ has $n-1$ maximum turning points.
  • Hence, to have at least three turning points, $f(x)$ must have a degree of $n \geq 4$.

Answer: $\text{Grad } n \geq 4$, da der Graph drei Wendepunkte besitzt.

b)

"Für jedes $x \in \mathbb{R}$ gilt $f(x) \geq -4$. Untersuchen Sie, ob auch $f'(x) \geq -4$ für jedes $x \in \mathbb{R}$ gilt."

• $f(x) \geq -4$ means $f(x)$'s minimum value is at least $-4$. The graph shows $f(x) = -4$ at its lowest points.
• However, $f'(x)$ represents the slope (derivative), and the slope can be less than $-4$.
  • Example: At steeply decreasing points, $f'(x) < -4$.
  • Checking the graph's slopes reveals regions where $f'(x)$ is negative and less than $-4$.

Answer: $f'(x) \geq -4$ gilt nicht, da der Graph von $f(x)$ steiler abfällt (z.B. $f'(x) < -4$ in bestimmten Intervallen).

c)

"F ist eine Stammfunktion von $f$. Untersuchen Sie, ob der Graph von $F$ mindestens einen Wendepunkt mit waagrechter Tangente besitzt."

• $F(x)$ is the antiderivative of $f(x)$, i.e., $F'(x) = f(x)$.
• A point of inflection (Wendepunkt) occurs when $F''(x) = f'(x)$ changes sign.
  • The graph of $f(x)$ shows regions where $f'(x)$ changes sign (e.g., at extrema).
  • At these points, $F''(x) = f'(x)$ crosses zero, confirming an inflection point.

Answer: Der Graph von $F(x)$ hat mindestens einen Wendepunkt mit waagrechter Tangente, da $f'(x)$ ein Vorzeichenwechsel hat.

d)

"Begründen Sie, dass gilt: $\int_0^2 f(x) dx > -8$."

• The definite integral $\int_0^2 f(x) dx$ represents the area under the curve of $f(x)$ from $x = 0$ to $x = 2$.
• The graph shows that $f(x) \geq -4$ everywhere, including this interval.
  • Since $f(x) \geq -4$, the total area under $f(x)$ is greater than $-8$ because the integral $\int_0^2 -4 dx = -8$.

Answer: $\int_0^2 f(x) dx > -8$, da $f(x) \geq -4$ in diesem Intervall.

e)

"Gegeben ist die Funktion $g(x) = f(x) + c$. Für einen Wert von $c$ gilt $\int_0^2 f(x) dx = 8 + \int_0^2 g(x) dx$. Ermitteln Sie den Wert von $c$."

• Start with the given equation:
  $\int_0^2 f(x) dx = 8 + \int_0^2 g(x) dx$
  Substitute $g(x) = f(x) + c$:
  $\int_0^2 f(x) dx = 8 + \int_0^2 \big(f(x) + c\big) dx$
  Expand the integral:
  $\int_0^2 f(x) dx = 8 + \int_0^2 f(x) dx + \int_0^2 c \, dx$
  Simplify:
  $0 = 8 + c \cdot 2 \quad \implies \quad c = -4$

Answer: ( c = -