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If $$\frac{3x^2-4x+1}{x-1}=m$$, and x can be any real number except 1, what real values can m NOT have?

Aug 1, 2019

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$$f(x) = \dfrac{3x^2-4x+1}{x-1}=\\ \dfrac{(3x-1 )(x-1 )}{x-1}=3x-1\\ \text{but with a "hole" at x=1 so 3(1)-1=2 is not in the range of f(x)}$$

$$\text{x = 1 is known as a removable singularity}$$

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Aug 1, 2019
edited by Rom  Aug 1, 2019