# How do you evaluate f(x+1) given the function f(x)=3-\frac{1}{2} x?

May 29, 2018

See a solution process below:

#### Explanation:

To evaluate $f \left(x + 1\right)$ substitute $\textcolor{red}{\left(x + 1\right)}$ for each occurrence of $\textcolor{red}{x}$ in $f \left(x\right)$ and calculate the result:

$f \left(\textcolor{red}{x}\right) = 3 - \frac{1}{2} \textcolor{red}{x}$ becomes:

$f \left(\textcolor{red}{x + 1}\right) = 3 - \frac{1}{2} \textcolor{red}{\left(x + 1\right)}$

$f \left(\textcolor{red}{x + 1}\right) = 3 - \frac{1}{2} x - \left(\frac{1}{2} \cdot 1\right)$

$f \left(\textcolor{red}{x + 1}\right) = 3 - \frac{1}{2} x - \frac{1}{2}$

$f \left(\textcolor{red}{x + 1}\right) = 3 - \frac{1}{2} - \frac{1}{2} x$

$f \left(\textcolor{red}{x + 1}\right) = \left(\frac{2}{2} \cdot 3\right) - \frac{1}{2} - \frac{1}{2} x$

$f \left(\textcolor{red}{x + 1}\right) = \frac{6}{2} - \frac{1}{2} - \frac{1}{2} x$

$f \left(\textcolor{red}{x + 1}\right) = \frac{6 - 1}{2} - \frac{1}{2} x$

$f \left(\textcolor{red}{x + 1}\right) = \frac{5}{2} - \frac{1}{2} x$

Or

$f \left(\textcolor{red}{x + 1}\right) = \frac{5 - x}{2}$