10. [-/0.83 Points] DETAILS MY NOTES SCALCET9 2.5.045. ASK YOUR TEACHER Find each \( x \)-value at which \( f \) is discontinuous and for each \( x \)-value, determine whether \( f \) is continuous from the right, or from the left, or netthen. \[ r(x)=\left\{\begin{array}{ll} x+6 & \text { if } x<0 \\ e^{x} & \text { if } 0 \leq x \leq 1 \\ 7-x & \text { if } x>1 \end{array}\right. \] \( x= \) \( \square \) (smaller value) Is \( f \) continuous from the right, left, or neither at this value? continuous from the right continuous from the left neither \( x= \) \( \square \) (larger value) Is \( f \) continuous from the right, left, or neither at this value? continuous from the right continuous from the left neither Sketch the graph of \( f \).

Let’s dive into the world of discontinuities! We'll start by identifying the points where the piecewise function \( r(x) \) has potential discontinuities. The breakpoints here are at \( x = 0 \) and \( x = 1 \). At \( x = 0 \): - From the left, \( r(0^-) = 0 + 6 = 6 \). - From the right, \( r(0^+) = e^{0} = 1 \). Since \( r(0^-) \neq r(0^+) \), \( r(x) \) is discontinuous at \( x = 0 \). Here, it's continuous from the left but neither from the right. At \( x = 1 \): - From the left, \( r(1^-) = e^{1} \approx 2.718 \). - From the right, \( r(1^+) = 7 - 1 = 6 \). Again, \( r(1^-) \neq r(1^+) \), so \( r(x) \) is discontinuous at \( x = 1 \). It is continuous from the left but neither from the right. Now let’s list our findings: 1. \( x = 0 \) - continuous from the left 2. \( x = 1 \) - continuous from the left Sketching the graph, you'd plot the pieces: a line from \( x < 0 \) going to \( y = 6 \), an exponential curve from \( x = 0 \) to \( x = 1 \) dipping down to \( y = 1 \) at \( x = 0 \), and a linear decline starting at \( y = 6 \) for \( x > 1 \). Be sure to mark those jumps at \( x = 0 \) and \( x = 1 \)!