Question

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\left\\{\begin{array}{ll} -2 x+3, & x<1 \\ x^{2}, & x \geq 1 \end{array}\right. $$

Short answer

The function is continuous for all x-values. There are no discontinuities.

Step-by-step solution

Understand the Definitions of Continuity and Removable Discontinuity

A function f is continuous at a point x=a if the following three conditions are satisfied: 1. \(f(a)\) is defined, 2. The limit as x approaches a of f(x) exists, 3. The limit as x approaches a of f(x) equals f(a). A discontinuity at x=a is removable if the function is undefined at x=a but the limit as x approaches a of f(x) exists.

Check Continuity on Each Piece of the Function

For x < 1, the function is -2x+3 which is a straight line and is continuous for all x. For x ≥ 1, the function is \(x^2\) which is a parabola and is also continuous for all x. Hence, there is no discontinuity in these intervals.

Check Continuity at the Junction Point

The junction point of these two pieces is at x=1. Calculate the limit of each piece at x=1. For x < 1, the limit as x approaches 1 is \(-2*1+3 = 1\). For x ≥ 1, the limit as x approaches 1 is \(1^2=1\). Both limits are equal, hence the function is continuous at x=1.

Conclusion

So, the function is continuous for all x. Therefore, there is no x-value at which the function is not continuous and no discontinuities exist, and the question of removable or non-removable discontinuities doesn't arise.