Calculus

Gegeben ist die Funktion f mit f(x) = 1/10 * (1/4 * x^4 - 9/2 * x^2) und der Definitionsmenge D_f = [-3; 4]. 1.1 Untersuchen Sie den Graphen G_f auf Symmetrie. 1.2 Bestimmen Sie die Schnittpunkte von G_f mit den Koordinatenachsen. 1.3 Ermitteln Sie das Monotonieverhalten der Funktion f. 1.4 Berechnen Sie die Koordinaten und die genaue Art aller Extrempunkte von G_f. Geben Sie auch die Wertemenge W_f an. 1.5 Zeichnen Sie den Graphen unter Verwendung aller bisherigen Ergebnisse in ein kartesisches Koordinatensystem.

Step-by-step solution with explanation

Final Answer

1.1 f is even → graph is symmetric about the y-axis (within the symmetric part of D_f). 1.2 Only intercept: origin S(0|0). 1.3 f increases on [−3, 0], decreases on [0, 3], increases on [3, 4]. 1.4 Local/global max: HP(0 | 0); local/global minima: TP(−3 | −81/40) and TP(3 | −81/40); right boundary value f(4) = −4/5. Range: W_f = [−81/40, 0] = [−2.025, 0].

Step-by-step solution

Simplify f(x) and check symmetry

We rewrite f(x) in a cleaner form. To check symmetry, we test if f(-x) = f(x) (even/y-axis symmetry) or f(-x) = -f(x) (odd/origin symmetry). Since f(-x) = \frac{(-x)^4}{40} - \frac{9(-x)^2}{20} = \frac{x^4}{40} - \frac{9x^2}{20} = f(x), the function is EVEN — its graph is symmetric about the y-axis. Note: the domain D_f = [-3, 4] is NOT symmetric about 0, so the graph as a whole is not perfectly symmetric, but the rule f(-x)=f(x) holds for all x where both sides are in the domain.

Find intercepts with coordinate axes

For x-intercepts we set f(x)=0 and factor. This gives x²=0 → x=0, or x²/40 = 9/20 → x² = 18 → x = ±√18 = ±3√2 ≈ ±4.24. But the domain is [-3, 4], so x = 3√2 ≈ 4.24 is outside and x = -3√2 ≈ -4.24 is also outside. Therefore the only x-intercept in D_f is x=0. The y-intercept is f(0)=0. So G_f meets both axes only at the origin: .

Find the first derivative f'(x)

We differentiate f(x) term by term. Factoring out x reveals three zeros of f'(x): x=0, x=3, and x=-3. These are the critical points (possible extrema). We check which lie in D_f=[-3,4]: all three do (x=-3 is a boundary point).

Determine monotone intervals using sign of f'(x)

We analyse the sign of f'(x) on D_f=[-3,4] by checking each interval between critical points. On (-3, 0): pick x=-1 → f'(-1)=(-1)(-4)(2)/10 = 8/10 > 0, so f is INCREASING. On (0, 3): pick x=1 → f'(1)=(1)(-2)(4)/10 = -8/10 < 0, so f is DECREASING. On (3, 4): pick x=3.5 → f'(3.5)=(3.5)(0.5)(6.5)/10 > 0, so f is INCREASING. Summary: f increases on [-3,0], decreases on [0,3], increases on [3,4].

Classify extrema and find their coordinates

At x=0, f' changes from + to −, so x=0 is a LOCAL MAXIMUM: . At x=3, f' changes from − to +, so x=3 is a LOCAL MINIMUM: . At x=-3 (left boundary), f is increasing to the right, so it is a boundary minimum: . We also compute the right boundary: .

State the range (Wertemenge) W_f

The global minimum value is , achieved at both x=-3 and x=3. The global maximum on D_f=[-3,4] is f(0)=0. All values between these are taken (the function is continuous), so the range is .

Understanding this problem

Learning Insight

A function is even when every term has an even exponent — flipping x to -x leaves every term unchanged because (-x)^{even} = x^{even}. That's why the graph mirrors perfectly across the y-axis. Factoring f'(x) completely is the key step: each sign change of f' at a critical point tells you exactly whether you have a hill (max) or a valley (min).

Quick Tip

To factor a polynomial derivative quickly, always pull out the common factor first (here: x), then recognise the difference-of-squares pattern x²-9=(x-3)(x+3). This instantly gives you all critical points without using the quadratic formula.

Common Mistake

Students often forget to check whether critical points actually lie inside the given domain. Here x=±3√2 are zeros of f but are outside D_f=[-3,4], so they are NOT intercepts — including them is a frequent error.