Question

sec 3.1 basics of functions: problem (1 point) given the function $f(x)=\frac{x + 1}{3x-3}$, calculate the following values: $f(0)=$ $f(-2)=$ $f(1)=$ note: if any of the values is not defined, type undefined. note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining.

Explanation:

Step1: Substitute \(x = 0\) into \(f(x)\)

Substitute \(x=0\) into \(f(x)=\frac{x + 1}{3x-3}\), we get \(f(0)=\frac{0 + 1}{3\times0-3}=\frac{1}{-3}=-\frac{1}{3}\).

Step2: Substitute \(x=-2\) into \(f(x)\)

Substitute \(x = - 2\) into \(f(x)=\frac{x + 1}{3x-3}\), we have \(f(-2)=\frac{-2 + 1}{3\times(-2)-3}=\frac{-1}{-6 - 3}=\frac{-1}{-9}=\frac{1}{9}\).

Step3: Substitute \(x = 1\) into \(f(x)\)

Substitute \(x = 1\) into \(f(x)=\frac{x + 1}{3x-3}\), the denominator \(3x-3=3\times1 - 3=0\). Since division by zero is undefined, \(f(1)\) is Undefined.

Answer:

\(f(0)=-\frac{1}{3}\)
\(f(-2)=\frac{1}{9}\)
\(f(1)=\text{Undefined}\)