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Suppose you're given the following table of values for the function f(x), and you're told that the function is even:x –2 –.35 0 .53 1f(x) 5 –3 2 2 –5Then:A.Something is wrong. Given the table of values, the function can't be even.B. f(0) + f(- .53) = 0C. f(- 1) -  f(2) = - 10D. f(.35) + f(- .53) = 1E. f(2) = - 5

Question

Suppose you're given the following table of values for the function f(x), and you're told that the function is even:x –2 –.35 0 .53 1f(x) 5 –3 2 2 –5Then:A.Something is wrong. Given the table of values, the function can't be even.B. f(0) + f(- .53) = 0C. f(- 1) -  f(2) = - 10D. f(.35) + f(- .53) = 1E. f(2) = - 5

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Solution

Para determinar cuál de las afirmaciones es correcta, primero recordemos que una función es par si y solo si f(x)=f(x) f(x) = f(-x) para todos los valores de x x en su dominio.

Dado el conjunto de valores:

x20.3500.531f(x)53225 \begin{array}{c|c|c|c|c|c} x & -2 & -0.35 & 0 & 0.53 & 1 \\ \hline f(x) & 5 & -3 & 2 & 2 & -5 \\ \end{array}

Verifiquemos cada afirmación:

A. Algo está mal. Dada la tabla de valores, la función no puede ser par.

  • Para que la función sea par, deberíamos tener f(x)=f(x) f(x) = f(-x) . Verifiquemos:
    • f(2)=5 f(-2) = 5 y f(2) f(2) no está dado, pero si asumimos que f(2)=5 f(2) = 5 , esto sería consistente.
    • f(0.35)=3 f(-0.35) = -3 y f(0.35) f(0.35) no está dado, pero si asumimos que f(0.35)=3 f(0.35) = -3 , esto sería consistente.
    • f(0)=2 f(0) = 2 , y f(0)=2 f(0) = 2 , esto es consistente.
    • f(0.53)=2 f(-0.53) = 2 y f(0.53)=2 f(0.53) = 2 , esto es consistente.
    • f(1) f(-1) no está dado, pero si asumimos que f(1)=5 f(-1) = -5 , esto sería consistente.
  • No hay contradicciones evidentes, por lo que la función podría ser par. Por lo tanto, la afirmación A es incorrecta.

B. f(0)+f(0.53)=0 f(0) + f(-0.53) = 0

  • f(0)=2 f(0) = 2 y f(0.53)=2 f(-0.53) = 2
  • 2+2=40 2 + 2 = 4 \neq 0
  • La afirmación B es incorrecta.

C. f(1)f(2)=10 f(-1) - f(2) = -10

  • f(1) f(-1) no está dado, pero si asumimos que f(1)=5 f(-1) = -5 (para que la función sea par), y f(2)=5 f(2) = 5
  • 55=10 -5 - 5 = -10
  • La afirmación C es correcta.

D. f(0.35)+f(0.53)=1 f(0.35) + f(-0.53) = 1

  • f(0.35) f(0.35) no está dado, pero si asumimos que f(0.35)=3 f(0.35) = -3 (para que la función sea par), y f(0.53)=2 f(-0.53) = 2
  • 3+2=11 -3 + 2 = -1 \neq 1
  • La afirmación D es incorrecta.

E. f(2)=5 f(2) = -5

  • No tenemos el valor de f(2) f(2) en la tabla, pero si asumimos que f(2)=5 f(2) = 5 (para que la función sea par), esto sería consistente.
  • La afirmación E es incorrecta.

Por lo tanto, la respuesta correcta es:

C. f(1)f(2)=10 f(-1) - f(2) = -10

This problem has been solved

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