Questions: f(x)=sqrt[2x](3x+1), x>0 sqrt(x), x=0

Solution Steps

To explore the continuity of the given piecewise function \( f(x) \), we need to check the limit of \( f(x) \) as \( x \) approaches 0 from the positive side and compare it with the value of \( f(x) \) at \( x = 0 \). If both are equal, the function is continuous at \( x = 0 \).

The piecewise function is defined as follows: \[ f(x) = \begin{cases} (3x + 1)^{\frac{1}{2x}}, & x > 0 \\ \sqrt{x}, & x = 0 \end{cases} \]

To check the continuity at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0 from the right: \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (3x + 1)^{\frac{1}{2x}} = e^{\frac{3}{2}} \]

Next, we evaluate the function at \( x = 0 \): \[ f(0) = \sqrt{0} = 0 \]

For the function to be continuous at \( x = 0 \), the limit from the right must equal the value of the function at that point: \[ e^{\frac{3}{2}} \neq 0 \] Since the limit does not equal the function value, the function is not continuous at \( x = 0 \).

Final Answer