Szkic wykresu f(x) = (x + 2)(x − 1)(x − 3) z postaci iloczynowej

Szkoła średnia łatwe

Naszkicuj wykres funkcji

$f(x) = (x + 2)(x - 1)(x - 3)$

i odczytaj z niego zbiór argumentów, dla których $f(x) > 0$ .

Rozwiązanie

Pokaż rozwiązanie krok po krokuUkryj rozwiązanie

Krok 1. Krańce wykresu. Wyraz najwyższego stopnia to $x \cdot x \cdot x = x^3$ : stopień $3$ (nieparzysty), współczynnik $a = 1 > 0$ . Prawy koniec wykresu idzie w górę, lewy w dół.

Krok 2. Miejsca zerowe. $x = -2$ , $x = 1$ , $x = 3$ - wszystkie pojedyncze, więc w każdym z nich wykres przecina oś $OX$ .

Krok 3. Punkt na osi $OY$ .

$f(0) = 2 \cdot (-1) \cdot (-3) = 6.$

Krok 4. Punkty pomocnicze. Po jednym między miejscami zerowymi:

$f(-1) = 1 \cdot (-2) \cdot (-4) = 8, \qquad f(2) = 4 \cdot 1 \cdot (-1) = -4.$

Krok 5. Szkic. Od prawej: wykres schodzi z góry, przecina oś w $3$ , schodzi pod oś (przez $(2,\ -4)$ ), wraca i przecina oś w $1$ , wznosi się nad oś (przez $(0,\ 6)$ i $(-1,\ 8)$ ), opada, przecina oś w $-2$ i dalej w lewo idzie w dół.

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Wykres f(x) = (x + 2)(x − 1)(x − 3): trzy pojedyncze miejsca zerowe −2, 1, 3 (wszędzie przecięcie osi), punkt (0, 6) na osi OY, lewy koniec w dole, prawy w górze.

Krok 6. Odczyt. Wykres leży nad osią $OX$ między $-2$ a $1$ oraz na prawo od $3$ . Nierówność jest ostra, więc miejsc zerowych nie dołączamy.

Odpowiedź. $f(x) > 0$ dla $x \in (-2;\ 1) \cup (3;\ +\infty)$ .

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