Question

1. find the discontinuities of

q(x) = \

$$\begin{cases} \\sqrt{\\cos(x) + 8}, & x > 0, \\\\ 3, & x = 0, \\\\ e^x, & x < 0. \\end{cases}$$

wherever q(x) is discontinuous, is it continuous from the left or right?

Explanation:

Step1: Recall the definition of continuity

A function \( y = q(x) \) is continuous at \( x = a \) if \( \lim_{x
ightarrow a^{-}}q(x)=\lim_{x
ightarrow a^{+}}q(x)=q(a) \). We need to check the continuity at \( x = 0 \) since the function is piece - wise defined with different expressions for \( x>0 \), \( x = 0 \) and \( x<0 \).

Step2: Calculate the left - hand limit as \( x

ightarrow0^{-} \)
For \( x<0 \), \( q(x)=e^{x} \). The left - hand limit \( \lim_{x
ightarrow0^{-}}q(x)=\lim_{x
ightarrow0^{-}}e^{x} \). Using the property of exponential functions, \( \lim_{x
ightarrow0^{-}}e^{x}=e^{0}=1 \).

Step3: Calculate the right - hand limit as \( x

ightarrow0^{+} \)
For \( x > 0 \), \( q(x)=\sqrt{\cos(x)+8} \). The right - hand limit \( \lim_{x
ightarrow0^{+}}q(x)=\lim_{x
ightarrow0^{+}}\sqrt{\cos(x)+8} \). Since \( \cos(0) = 1 \), we have \( \lim_{x
ightarrow0^{+}}\sqrt{\cos(x)+8}=\sqrt{\cos(0)+8}=\sqrt{1 + 8}=\sqrt{9}=3 \).

Step4: Find the value of the function at \( x = 0 \)

Given that \( q(0)=3 \).

Step5: Analyze continuity at \( x = 0 \)

We have \( \lim_{x
ightarrow0^{-}}q(x)=1 \), \( \lim_{x
ightarrow0^{+}}q(x)=3 \) and \( q(0) = 3 \). Since \( \lim_{x
ightarrow0^{-}}q(x)
eq\lim_{x
ightarrow0^{+}}q(x) \), the function is discontinuous at \( x = 0 \). Now, check the right - hand continuity: \( \lim_{x
ightarrow0^{+}}q(x)=q(0)=3 \), so the function is continuous from the right at \( x = 0 \).

Answer:

The function \( q(x) \) is discontinuous at \( x = 0 \) and it is continuous from the right at \( x = 0 \).