Let f be the function defined as \lfloor x-\dfrac{3}{5} \rfloor. What is f\left( \dfrac{3}{7} \right) ?
The greatest integer function \lfloor a\rfloor determines the greatest integer less than or equal to a.
In our problem:
f\left( \dfrac{3}{7} \right)=\lfloor \dfrac{3}{7}-\dfrac{3}{5} \rfloor=\lfloor -\dfrac{6}{35} \rfloor
Since -\dfrac{6}{35}\in\left[-1{,}0\right), we can conclude that:
f\left( \dfrac{3}{7} \right)=-1
Let f be the function defined as \lfloor x+\dfrac{1}{2} \rfloor. What is f\left( \dfrac{3}{4} \right) ?
The greatest integer function \lfloor a\rfloor determines the greatest integer less than or equal to a.
In our problem:
f\left( \dfrac{3}{4} \right)=\lfloor\dfrac{3}{4}+\dfrac{1}{2} \rfloor=\lfloor \dfrac{5}{4} \rfloor
Since \dfrac{5}{4}\in\left[1{,}2\right), we can conclude that:
f\left( \dfrac{3}{4} \right)=1
Let f be the function defined as \lfloor 3x-\dfrac{1}{4} \rfloor. What is f\left( -\dfrac{1}{2} \right) ?
The greatest integer function \lfloor a\rfloor determines the greatest integer less than or equal to a.
In our problem:
f\left( -\dfrac{1}{2} \right)=\lfloor 3\cdot\left( -\dfrac{1}{2} \right)-\dfrac{1}{4} \rfloor=\lfloor -\dfrac{7}{4} \rfloor
Since -\dfrac{7}{4}\in\left[-2,-1\right), we can conclude that:
f\left( -\dfrac{1}{2} \right)=-2
Let f be the function defined as \lfloor 5x+\dfrac{2}{3} \rfloor. What is f\left( -\dfrac{1}{3} \right) ?
The greatest integer function \lfloor a\rfloor determines the greatest integer less than or equal to a.
In our problem:
f\left( -\dfrac{1}{3} \right)=\lfloor5\cdot\left( -\dfrac{1}{3} \right)+\dfrac{2}{3} \rfloor=\lfloor -1 \rfloor
Since -1\in\left[-1{,}0\right), we can conclude that:
f\left( -\dfrac{1}{3} \right)=-1
Let f be the function defined as \lfloor x-\dfrac{7}{5} \rfloor. What is f\left( \dfrac{3}{10} \right) ?
The greatest integer function \lfloor a\rfloor determines the greatest integer less than or equal to a.
In our problem:
f\left( \dfrac{3}{10} \right)=\lfloor \dfrac{3}{10}-\dfrac{7}{5} \rfloor=\lfloor -\dfrac{11}{10} \rfloor
Since -\dfrac{11}{10}\in\left[-2,-1\right), we can conclude that:
f\left( \dfrac{3}{10} \right)=-2
Let f be the function defined as \lfloor x-\dfrac{2}{3} \rfloor . What is f\left( \dfrac{5}{6} \right) ?
The greatest integer function \lfloor a\rfloor determines the greatest integer less than or equal to a.
In our problem:
f\left( \dfrac{5}{6} \right)=\lfloor \dfrac{5}{6}-\dfrac{2}{3} \rfloor=\lfloor \dfrac{1}{6} \rfloor
Since \dfrac{1}{6}\in\left[0{,}1\right), we can conclude that:
f\left( \dfrac{5}{6} \right)=0
Let f be the function defined as \lfloor-3x+\dfrac{1}{4} \rfloor . What is f\left( \dfrac{3}{8} \right) ?
The greatest integer function \lfloor a\rfloor determines the greatest integer less than or equal to a.
In our problem:
f\left( \dfrac{3}{8} \right)=\lfloor -3\left( \dfrac{3}{8}\right)+\dfrac{1}{4} \rfloor=\lfloor -\dfrac{9}{8}+\dfrac{1}{4} \rfloor=\lfloor -\dfrac{7}{8} \rfloor
Since -\dfrac{7}{8}\in\left[-1{,}0\right), we can conclude that:
f\left( \dfrac{3}{8} \right)=-1