Question

Find the value of $f\left(2\right),f\left(-2\right),f\left(\frac{1}{2}\right),f\left(a\right),f\left(-a\right),f\left(a-1\right)$for the function $f\left(x\right)=\frac{1-2x}{3}$.

Short answer

The value of function are

$f\left(2\right)=-1,f\left(-2\right)=\frac{5}{3},f\left(\frac{1}{2}\right)=0,f\left(a\right)=\frac{1-2a}{3},f\left(-a\right)=\frac{1+2a}{3},f\left(a-1\right)=\frac{3-2a}{3}$.

Step-by-step solution

Step 1. Evaluate f2,f−2,f12 for the given function.

From the given function,

$f\left(x\right)=\frac{1-2x}{3}\mathrm{....}\left(1\right)$

Evaluate the value of function at $x=2,-2,\frac{1}{2}$gives,

role="math" localid="1644495962389" $$\begin{gathered} f\left(2\right)=\frac{1-2\left(2\right)}{3}\mathrm{....}\left(\text{Substitute}x=2\text{in}\left(1\right)\right) \\ =-1 \\ f\left(-2\right)=\frac{1-2\left(-2\right)}{3}\mathrm{....}\left(\text{Substitute}x=-2\text{\hspace{0.17em}in}\left(1\right)\right) \\ =\frac{5}{3} \\ f\left(\frac{1}{2}\right)=\frac{1-2\left(\frac{1}{2}\right)}{3}....\left(\text{Substitute}x=\frac{1}{2}\text{in}\left(1\right)\right) \\ =0 \end{gathered}$$

Step 2. Evaluate fa,f−a,fa−1 for the given function.

Similarly, evaluate the value of function (1) at $x=a,-a,a-1$gives,

role="math" localid="1644496293794" $$\begin{gathered} f\left(a\right)=\frac{1-2\left(a\right)}{3}\mathrm{....}\left(\text{Substitute}x=a\text{in}\left(1\right)\right) \\ =\frac{1-2a}{3} \\ f\left(-a\right)=\frac{1-2\left(-a\right)}{3}\mathrm{....}\left(\text{Substitute}x=-a\text{\hspace{0.17em}in}\left(1\right)\right) \\ =\frac{1+2a}{3} \\ f\left(a-1\right)=\frac{1-2\left(a-1\right)}{3}\mathrm{....}\left(\text{Substitute}x=a-1\text{in}\left(1\right)\right) \\ =\frac{3-2a}{3} \end{gathered}$$

Step 3. Description of steps.

From the obtained values

$f\left(2\right)=-1,f\left(-2\right)=\frac{5}{3},f\left(\frac{1}{2}\right)=0,f\left(a\right)=\frac{1-2a}{3},f\left(-a\right)=\frac{1+2a}{3},f\left(a-1\right)=\frac{3-2a}{3}$.