find each x - value at which f is discontinuous and for each x - value, determine whether f is continuous…

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 n\n f(x)=\begin{cases}4^{x}&\text{if }xleq1\\5 - x&\text{if }1lt xleq6\\sqrt{x}&\text{if }xgt6end{cases} \nis f continuous from the right, left, or neither at this value?\ncontinuous from the right\ncontinuous from the left\nneither\nsketch the graph of f

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 n\n f(x)=\begin{cases}4^{x}&\text{if }xleq1\\5 - x&\text{if }1lt xleq6\\sqrt{x}&\text{if }xgt6end{cases} \nis f continuous from the right, left, or neither at this value?\ncontinuous from the right\ncontinuous from the left\nneither\nsketch the graph of f

Answer

Explanation:

Step1: Check continuity at (x = 1)

  • Left - hand limit: (\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{-}}4^{x}=4^{1} = 4)
  • Right - hand limit: (\lim_{x\rightarrow1^{+}}f(x)=\lim_{x\rightarrow1^{+}}(5 - x)=5-1 = 4)
  • Function value: (f(1)=4^{1}=4) Since (\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{+}}f(x)=f(1)), the function is continuous at (x = 1)

Step2: Check continuity at (x = 6)

  • Left - hand limit: (\lim_{x\rightarrow6^{-}}f(x)=\lim_{x\rightarrow6^{-}}(5 - x)=5 - 6=-1)
  • Right - hand limit: (\lim_{x\rightarrow6^{+}}f(x)=\lim_{x\rightarrow6^{+}}\sqrt{x}=\sqrt{6}\approx2.45)
  • Function value: (f(6)=5 - 6=-1)

Since (\lim_{x\rightarrow6^{-}}f(x)=f(6)=-1) and (\lim_{x\rightarrow6^{-}}f(x)\neq\lim_{x\rightarrow6^{+}}f(x))

Answer:

(x = 6), continuous from the left.