Question Statement

For the function $f(x)=\frac{1}{\sqrt{x}}$, verify that $f(x)$ is continuous on the following intervals:

(a) $[1,4]$

(b) $[1,9]$

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Background and Explanation

This problem requires checking the continuity of a radical function on closed intervals. A function is continuous on an interval if it is defined at every point within that interval and has no breaks, jumps, or asymptotes.

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Solution

Part (a): Interval $[1,4]$

The function $f(x)=\frac{1}{\sqrt{x}}$ is defined for all $x\in [1,4]$.

Since $f(x)$ is defined at every point in the closed interval $[1,4]$, and since $f(x)$ is continuous on its domain $(0,\infty )$, we conclude that $f(x)$ is continuous on $[1,4]$.

Part (b): Interval $[1,9]$

Similarly, $f(x)=\frac{1}{\sqrt{x}}$ is defined for all $x\in [1,9]$.

Therefore, $f(x)$ is continuous on the interval $[1,9]$.

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Key Formulas or Methods Used

• Domain of radical functions: $f(x)=\frac{1}{\sqrt{x}}$ requires $x>0$ to be defined
• Continuity on closed intervals: If a function is defined at every point in $[a,b]$ and is continuous on its domain, then it is continuous on $[a,b]$

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Summary of Steps

1. Identify that $f(x)=\frac{1}{\sqrt{x}}$ requires $x>0$ for the square root to be defined and non-zero
2. Verify that all $x\in [1,4]$ satisfy $x>0$, confirming $f(x)$ is defined on $[1,4]$
3. Conclude continuity on $[1,4]$ since the function is defined throughout the interval
4. Verify that all $x\in [1,9]$ satisfy $x>0$, confirming $f(x)$ is defined on $[1,9]$
5. Conclude continuity on $[1,9]$ since the function is defined throughout the interval