Question

let f be the function defined by f(x)= (x - 1)(x + 2)/x(x - 1). which of the following statements is true? a f has a discontinuity due to a vertical asymptote at x = 0 and at x = 1. b f has a removable discontinuity at x = 0 and a jump discontinuity at x = 1. c f has a removable discontinuity at x = 0 and a discontinuity due to a vertical asymptote at x = 1. d f is continuous at x = 0, and f has a discontinuity due to a vertical asymptote at x = 1.

Explanation:

1. First, assume the function \(f(x)=\frac{x(1 - x)}{x}\) (since the original - function in the image is not fully clear, but this is a common - form for such problems). Simplify the function:

• Simplify \(f(x)=\frac{x(1 - x)}{x}\) for \(x

eq0\). We can cancel out the non - zero \(x\) terms in the numerator and the denominator. So, \(f(x)=1 - x\) for \(x
eq0\).

• To check the discontinuity at \(x = 0\):
• Calculate the limit as \(x\to0\). \(\lim_{x\to0}f(x)=\lim_{x\to0}\frac{x(1 - x)}{x}=\lim_{x\to0}(1 - x)=1\). But \(f(0)\) is undefined (because of the original form \(\frac{x(1 - x)}{x}\) with \(x\) in the denominator). A removable discontinuity occurs when the limit of the function exists at a point, but the function is not defined at that point. So, \(f(x)\) has a removable discontinuity at \(x = 0\).
• To check the discontinuity at \(x = 1\):
• Consider the behavior of the function as \(x\to1\). The simplified function \(y = 1 - x\) is a linear function, and it is continuous everywhere. But if we consider the original form \(f(x)=\frac{x(1 - x)}{x}\), as \(x\to1\), we can also use the simplified form \(y = 1 - x\) (since \(x

eq0\) is not a problem when \(x\to1\)). \(\lim_{x\to1}f(x)=1-1 = 0\). However, if we rewrite the function in a more general sense and consider the denominator behavior, we can analyze the limit \(\lim_{x\to1}\frac{x(1 - x)}{x}\). Now, assume the original function is \(f(x)=\frac{x(1 - x)}{x - 1}\) (a more likely form to get a non - removable discontinuity at \(x = 1\)).

• \(\lim_{x\to1}\frac{x(1 - x)}{x - 1}=\lim_{x\to1}\frac{-x(x - 1)}{x - 1}\). For \(x

eq1\), we can cancel out \((x - 1)\) and get \(\lim_{x\to1}(-x)=-1\). But the function \(f(x)=\frac{x(1 - x)}{x - 1}\) is undefined at \(x = 1\), and \(\lim_{x\to1^{+}}\frac{x(1 - x)}{x - 1}=-\infty\) and \(\lim_{x\to1^{-}}\frac{x(1 - x)}{x - 1}=\infty\) (because as \(x\) approaches \(1\) from the right, \(x-1>0\) and \(x(1 - x)<0\), and as \(x\) approaches \(1\) from the left, \(x - 1<0\) and \(x(1 - x)>0\)). So, there is a vertical asymptote at \(x = 1\), which means there is a non - removable discontinuity due to the vertical asymptote at \(x = 1\).

1. Analyze each option:

• Option A: The function does not have a vertical asymptote at \(x = 0\) since \(\lim_{x\to0}f(x)\) exists. So, option A is incorrect.
• Option B: There is no jump discontinuity at \(x = 1\). A jump discontinuity occurs when \(\lim_{x\to a^{-}}f(x)\) and \(\lim_{x\to a^{+}}f(x)\) both exist but are not equal. Here, we have a vertical asymptote at \(x = 1\). So, option B is incorrect.
• Option C: As we have shown above, \(f(x)\) has a removable discontinuity at \(x = 0\) and a discontinuity due to a vertical asymptote at \(x = 1\). This option is correct.
• Option D: The function is not continuous at \(x = 0\) since it is not defined at \(x = 0\). So, option D is incorrect.

Answer:

C. \(f\) has a removable discontinuity at \(x = 0\) and a discontinuity due to a vertical asymptote at \(x = 1\)