For the function given below, decide whether \( f \) is continuous at 1 . If \( f \) is not continuous at 1 , classify the discontinuity as removable, jump, or infinite. \[ f(x)=\left\{\begin{array}{lll} x^{2} & \text { if } & x \neq 1 \\ 6 & \text { if } & x=1 \end{array}\right. \] \( f(x) \) is continuous at \( x=1 \). There is a removable discontinuity at \( \boldsymbol{x}=\mathbf{1} \). There is a jump discontinuity at \( \boldsymbol{x}=\mathbf{1} \). There is an infinite discontinuity at \( \boldsymbol{x}=\mathbf{1} \).

To determine the continuity of the given function \( f \) at \( x = 1 \), we first evaluate \( \lim_{x \to 1} f(x) \). As \( x \) approaches 1, \( f(x) = x^2 \) (for \( x \neq 1 \)), which approaches \( 1^2 = 1 \). However, \( f(1) = 6 \). Since \( \lim_{x \to 1} f(x) = 1 \) and \( f(1) = 6 \), the limit and the function value do not match. This means \( f \) is not continuous at \( x = 1 \). Moreover, because the limit exists but does not equal the function's value, we classify this as a removable discontinuity at \( x = 1 \). Another way to think about it is that we can redefine \( f(1) \) to be 1 instead of 6—like fixing a little "oops" in the function. If we made that change, the function would become continuous at that point!