Question

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\frac{x-1}{x^{2}+x-2} $$

Short answer

The function \( f(x) \) is discontinuous at \( x = 1 \) and \( x = -2 \). The discontinuity at \( x = 1 \) is removable while the discontinuity at \( x = -2 \) is not removable.

Step-by-step solution

Identify the Points of Discontinuity

Discontinuity occurs when the denominator of a fraction is zero because you can’t divide by zero. So, set the denominator equal to zero and solve for \( x \). This gives the equation \( x^{2} + x - 2 = 0 \).

Factor the Equation to Solve for \( x \)

The factored form of the equation is \( (x-1)(x+2) = 0 \). To find the \( x \)-values, set each factor equal to zero and solve. This gives \( x = 1 \) and \( x = -2 \)

Identify Removable Discontinuities

For removable discontinuity, the factors causing zero in the denominator should be cancelled out by the numerator. In the function \( f(x)=\frac{x-1}{x^{2}+x-2} \), after factoring, we get \( f(x)=\frac{x-1}{(x-1)(x+2)} \). Here, \( (x-1) \) can be cancelled from the numerator and the denominator. After canceling \( (x-1) \), we get \( f(x)=\frac{1}{x+2} \), which is defined at \( x=1 \). Hence, the discontinuity at \( x=1 \) is removable.

Final Answer

So, the function \( f(x) \) is discontinuous at \( x = 1 \) and \( x = -2 \). The discontinuity at \( x = 1 \) is removable while the discontinuity at \( x = -2 \) is not removable.