Bond Kirk
07/10/2023 · High School
En los problemas siguientes, calcule el valor de la función cuando \( x=-1, x=1 / 2 \) y \( x=0,8 \) a) \( f(x)=\frac{\lceil x\rceil}{x} \) b) \( f(x)=\frac{\left\lceil x-\frac{x}{3}\right\rceil}{x} \) c) \( f(x)=\frac{\lfloor x]}{x} \) d) \( f(x)=\frac{x}{x+\frac{1}{2}} \).
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Step-by-step Solution
Vamos a calcular el valor de las funciones dadas para los valores \( x = -1 \), \( x = \frac{1}{2} \) y \( x = 0.8 \).
### a) \( f(x) = \frac{\lceil x \rceil}{x} \)
- **Para \( x = -1 \)**:
\[
\lceil -1 \rceil = -1 \quad \Rightarrow \quad f(-1) = \frac{-1}{-1} = 1
\]
- **Para \( x = \frac{1}{2} \)**:
\[
\lceil \frac{1}{2} \rceil = 1 \quad \Rightarrow \quad f\left(\frac{1}{2}\right) = \frac{1}{\frac{1}{2}} = 2
\]
- **Para \( x = 0.8 \)**:
\[
\lceil 0.8 \rceil = 1 \quad \Rightarrow \quad f(0.8) = \frac{1}{0.8} = 1.25
\]
### b) \( f(x) = \frac{\left\lceil x - \frac{x}{3} \right\rceil}{x} \)
Primero simplificamos \( x - \frac{x}{3} = \frac{2x}{3} \).
- **Para \( x = -1 \)**:
\[
\left\lceil -1 - \frac{-1}{3} \right\rceil = \left\lceil -1 + \frac{1}{3} \right\rceil = \left\lceil -\frac{2}{3} \right\rceil = -1 \quad \Rightarrow \quad f(-1) = \frac{-1}{-1} = 1
\]
- **Para \( x = \frac{1}{2} \)**:
\[
\left\lceil \frac{2 \cdot \frac{1}{2}}{3} \right\rceil = \left\lceil \frac{1}{3} \right\rceil = 1 \quad \Rightarrow \quad f\left(\frac{1}{2}\right) = \frac{1}{\frac{1}{2}} = 2
\]
- **Para \( x = 0.8 \)**:
\[
\left\lceil \frac{2 \cdot 0.8}{3} \right\rceil = \left\lceil \frac{1.6}{3} \right\rceil = \left\lceil \frac{8}{15} \right\rceil = 1 \quad \Rightarrow \quad f(0.8) = \frac{1}{0.8} = 1.25
\]
### c) \( f(x) = \frac{\lfloor x \rfloor}{x} \)
- **Para \( x = -1 \)**:
\[
\lfloor -1 \rfloor = -1 \quad \Rightarrow \quad f(-1) = \frac{-1}{-1} = 1
\]
- **Para \( x = \frac{1}{2} \)**:
\[
\lfloor \frac{1}{2} \rfloor = 0 \quad \Rightarrow \quad f\left(\frac{1}{2}\right) = \frac{0}{\frac{1}{2}} = 0
\]
- **Para \( x = 0.8 \)**:
\[
\lfloor 0.8 \rfloor = 0 \quad \Rightarrow \quad f(0.8) = \frac{0}{0.8} = 0
\]
### d) \( f(x) = \frac{x}{x + \frac{1}{2}} \)
- **Para \( x = -1 \)**:
\[
f(-1) = \frac{-1}{-1 + \frac{1}{2}} = \frac{-1}{-0.5} = 2
\]
- **Para \( x = \frac{1}{2} \)**:
\[
f\left(\frac{1}{2}\right) = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{2}} = \frac{\frac{1}{2}}{1} = \frac{1}{2}
\]
- **Para \( x
Quick Answer
### a) \( f(x) = \frac{\lceil x \rceil}{x} \)
- **Para \( x = -1 \)**: \( f(-1) = 1 \)
- **Para \( x = \frac{1}{2} \)**: \( f\left(\frac{1}{2}\right) = 2 \)
- **Para \( x = 0.8 \)**: \( f(0.8) = 1.25 \)
### b) \( f(x) = \frac{\left\lceil x - \frac{x}{3} \right\rceil}{x} \)
- **Para \( x = -1 \)**: \( f(-1) = 1 \)
- **Para \( x = \frac{1}{2} \)**: \( f\left(\frac{1}{2}\right) = 2 \)
- **Para \( x = 0.8 \)**: \( f(0.8) = 1.25 \)
### c) \( f(x) = \frac{\lfloor x \rfloor}{x} \)
- **Para \( x = -1 \)**: \( f(-1) = 1 \)
- **Para \( x = \frac{1}{2} \)**: \( f\left(\frac{1}{2}\right) = 0 \)
- **Para \( x = 0.8 \)**: \( f(0.8) = 0 \)
### d) \( f(x) = \frac{x}{x + \frac{1}{2}} \)
- **Para \( x = -1 \)**: \( f(-1) = 2 \)
- **Para \( x = \frac{1}{2} \)**: \( f\left(\frac{1}{2}\right) = \frac{1}{2} \)
- **Para \( x = 0.8 \)**: \( f(0.8) = \frac{8}{10} = 0.8 \)
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