Question

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
f(x)=\begin{cases}4^{x}&\text{if }xleq1\\5 - x&\text{if }1lt xleq6\sqrt{x}&\text{if }xgt6end{cases}
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

Explanation:

Step1: Check continuity at \(x = 1\)

• Left - hand limit: \(\lim_{x

ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}4^{x}=4^{1} = 4\)

• Right - hand limit: \(\lim_{x

ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(5 - x)=5-1 = 4\)

• Function value: \(f(1)=4^{1}=4\)

Since \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)=f(1)\), the function is continuous at \(x = 1\)

Step2: Check continuity at \(x = 6\)

• Left - hand limit: \(\lim_{x

ightarrow6^{-}}f(x)=\lim_{x
ightarrow6^{-}}(5 - x)=5 - 6=-1\)

• Right - hand limit: \(\lim_{x

ightarrow6^{+}}f(x)=\lim_{x
ightarrow6^{+}}\sqrt{x}=\sqrt{6}\approx2.45\)

• Function value: \(f(6)=5 - 6=-1\)

Since \(\lim_{x
ightarrow6^{-}}f(x)=f(6)=-1\) and \(\lim_{x
ightarrow6^{-}}f(x)
eq\lim_{x
ightarrow6^{+}}f(x)\)

Answer:

\(x = 6\), continuous from the left.