Question

simplify f(x) = 3x^2-4x+1/3x-1
Part 2: Using complete sentences, explain why f(1) = 0, f(0) = -1, and f(-1) = -2, yet f(one-third) is undefined. Be sure to show your work. (4 points)

Answer 1

\[f(x)=\frac{3x^2-4x+1}{3x-1}\] right?

Answer 2

First of all, I think you forgot the parentheses and the actual equation is (3x^2-4x+1)/(3x-1). Make sure to put parentheses because you have to follow the order of parentheses, exponents, multiply/divide, and add/subtract when solving an equation.

f(1) = 0 because if you plug in x=1 for the equation f(x)=(3x^2-4x+1)/(3x-1) it becomes f(1)=(3(1)^2-4(1)+1)/(3(1)-1) which becomes (3-4+1)/(3-1) which becomes 0/3 which is 0.

f(0) = -1 because if you substitute in x=0 into the equation, it becomes (3(0)^2-4(0)+1)/(0-1) which becomes 1/-1 which is -1

f(-1) = -2 because once again if you plug x=-1 into the original equation you get (3(-1)^2-4(-1)+1)/3(-1)-1 which becomes (3+4+1)/(-3-1) which becomes 8/-4 which is -2

For this one I think you meant to write that f(1/3) is undefined because if x=1/3, then the denominator of the equation (3x-1) becomes zero and it is impossible to divide and number by zero. Therefore f(1/3) is undefined

Answer 3

right

Answer 4

\(f(1)=0\) because if you replace \(x\) by \(0\) you get \[f(x)=\frac{3\times 1^2-4\times 1+1}{3\times 1-1}=\frac{0}{2}=0\]

Answer 5

similarly \(f(-1)=0\)

Answer 6

oops i meant to write \[f(1)=\frac{3\times 1^2-4\times 1+1}{3\times 1-1}=\frac{0}{2}=0\]

Answer 7

and \(f(\frac{1}{3})\) is undefined, because the denominator would be \(3\times \frac{1}{3}-1=1-1=0\) and you cannot divide by \(0\)

Answer 8

Thanks