The function \( f(x) = rac{2x^2 - 3x + 1}{x - 1} \) has a removable discontinuity. Find the value of \( f(1) \) after removing the discontinuity.
["Understanding the Function ( f(x) = \frac{2x^2 - 3x + 1}{x - 1} ) and Its Removable Discontinuity", "When analyzing rational functions like ( f(x) = \frac{2x^2 - 3x + 1}{x - 1} ), it’s common to encounter discontinuities that can be removed through simplification. In this article, we explore the behavior of ( f(x) ), identify its removable discontinuity at ( x = 1 ), and determine the meaningful value of ( f(1) ) after safely removing the discontinuity.", "---", "### What Is a Removable Discontinuity?", "A removable discontinuity occurs at a point ( x = a ) where the function is undefined or appears to “blow up,” but the limit exists. This means the function has a hole at ( x = a ) rather than a vertical asymptote. Such discontinuities arise when the numerator and denominator share a common factor, allowing the expression to be simplified.", "---", "### Analyzing the Function ( f(x) = \frac{2x^2 - 3x + 1}{x - 1} )", "Start by studying the numerator:", "[\n2x^2 - 3x + 1\n]", "Factor the quadratic expression:", "[\n2x^2 - 3x + 1 = (2x - 1)(x - 1)\n]", "Thus, the function becomes:", "[\nf(x) = \frac{(2x - 1)(x - 1)}{x - 1}\n]", "For all ( x <br/>\neq 1 ), we can simplify:", "[\nf(x) = 2x - 1\n]", "However, at ( x = 1 ), the original expression is undefined because the denominator becomes zero while the numerator is not zero:", "[\n2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0\n]", "So, ( f(1) ) is undefined in the original form due to division by zero—yet the numerator also equals zero, confirming a 0/0 indeterminate form. This signals a potential removable discontinuity at ( x = 1 ).", "---", "### Identifying and Removing the Discontinuity", "Even though both numerator and denominator vanish at ( x = 1 ), simplifying the function for ( x <br/>\ne 1 ) gives:", "[\nf(x) = 2x - 1\n]", "By removing the common factor ( (x - 1) ), the discontinuity is removable. To define ( f(1) ) continuously, replace ( f(1) ) with the limit as ( x \ o 1 ):", "[\n\lim_{x \ o 1} f(x) = \lim_{x \ o 1} (2x - 1) = 2(1) - 1 = 1\n]", "Thus, defining ( f(1) = 1 ) removes the discontinuity and makes ( f(x) ) continuous at ( x = 1 ).", "---", "### Conclusion", "The function ( f(x) = \frac{2x^2 - 3x + 1}{x - 1} ) has a removable discontinuity at ( x = 1 ) because the numerator and denominator share a common factor, allowing simplification to ( 2x - 1 ), with a hole at ( x = 1 ). By evaluating the limit, we find that the value ( f(1) ) should be defined as 1 to eliminate the discontinuity.", "This approach exemplifies how factoring and limit analysis can resolve apparent breaks in rational functions, enabling clean, meaningful function definitions across all real numbers.", "---", "### Practical Takeaway", "When faced with a rational function where ( f(1) ) appears undefined, check whether numerator and denominator share common roots. If so, simplify and compute the limit at that point to define a value that removes the discontinuity. For this function,\n[\n\boxed{f(1) = 1}\n]\nmakes ( f(x) ) continuous on ( \mathbb{R} )."]
