Stanley Wagner
05/13/2024 · Middle School
a) Para la función \( f(x)=x^{2}+2 x+1 \) coloca los símbolos \( <,>0= \) donde corresponda. - \( f(-2) \quad f(0) \) - \( f(-3) \quad f(-1) \) - \( f(1) \quad f(3) \)
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Tutor-Verified Answer
Step-by-step Solution
Primero, evaluemos la función \( f(x) = x^2 + 2x + 1 \) en los puntos dados.
La función se puede reescribir como \( f(x) = (x + 1)^2 \), lo que nos indica que siempre será mayor o igual a cero, ya que el cuadrado de cualquier número es no negativo.
Ahora evaluemos los puntos:
1. **Para \( f(-2) \)**:
\[
f(-2) = (-2 + 1)^2 = (-1)^2 = 1 > 0
\]
2. **Para \( f(0) \)**:
\[
f(0) = (0 + 1)^2 = 1^2 = 1 > 0
\]
Por lo tanto, tenemos:
\[
f(-2) > 0 \quad f(0) > 0
\]
3. **Para \( f(-3) \)**:
\[
f(-3) = (-3 + 1)^2 = (-2)^2 = 4 > 0
\]
4. **Para \( f(-1) \)**:
\[
f(-1) = (-1 + 1)^2 = 0^2 = 0
\]
Por lo tanto, tenemos:
\[
f(-3) > 0 \quad f(-1) = 0
\]
5. **Para \( f(1) \)**:
\[
f(1) = (1 + 1)^2 = 2^2 = 4 > 0
\]
6. **Para \( f(3) \)**:
\[
f(3) = (3 + 1)^2 = 4^2 = 16 > 0
\]
Por lo tanto, tenemos:
\[
f(1) > 0 \quad f(3) > 0
\]
Resumiendo los resultados:
- \( f(-2) > 0 \quad f(0) > 0 \)
- \( f(-3) > 0 \quad f(-1) = 0 \)
- \( f(1) > 0 \quad f(3) > 0 \)
Así que los símbolos quedan de la siguiente manera:
- \( f(-2) > 0 \quad f(0) > 0 \)
- \( f(-3) > 0 \quad f(-1) = 0 \)
- \( f(1) > 0 \quad f(3) > 0 \)
Quick Answer
- \( f(-2) > 0 \quad f(0) > 0 \)
- \( f(-3) > 0 \quad f(-1) = 0 \)
- \( f(1) > 0 \quad f(3) > 0 \)
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