Question Statement

Determine whether the function $f(x)=\sqrt{x^2-9}$ is continuous on the following intervals:

(a) $[-3,3]$

(b) $[3,\infty )$

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Background and Explanation

For a function to be continuous on an interval, it must first be defined at every point within that interval. The square root function $\sqrt{u}$ is only defined for non-negative real numbers ($u\ge 0$), and compositions of continuous functions remain continuous on their domains.

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Solution

Part (a): Interval $[-3,3]$

To analyze continuity, we first determine where $f(x)$ is defined. The expression under the square root must satisfy:

$x^2-9\ge 0$

$x^2\ge 9$

$|x|\ge 3$

This means the domain of $f(x)$ is $(-\infty ,-3]\cup [3,\infty )$.

Now consider the open subinterval $(-3,3)\subset [-3,3]$. For any $x\in (-3,3)$:

• We have $x^2<9$
• Therefore $x^2-9<0$

When $x^2-9<0$, the value $\sqrt{x^2-9}$ is imaginary (undefined in the real number system). Thus, $f(x)$ is not defined for any $x\in (-3,3)$.

Since $f(x)$ is undefined throughout the interior of the interval $[-3,3]$, it cannot be continuous on $[-3,3]$.

Conclusion: $f(x)$ is discontinuous on $[-3,3]$.

Part (b): Interval $[3,\infty )$

For all $x\ge 3$:

• We have $x^2\ge 9$
• Therefore $x^2-9\ge 0$

This ensures that $f(x)=\sqrt{x^2-9}$ is defined for every $x\in [3,\infty )$.

Furthermore, $f(x)$ is a composition of two continuous functions:

1. The polynomial $g(x)=x^2-9$, which is continuous for all real $x$
2. The square root function $h(u)=\sqrt{u}$, which is continuous for all $u\ge 0$

Since $x^2-9\ge 0$ for all $x\in [3,\infty )$, the composition $f(x)=h(g(x))$ is continuous on $[3,\infty )$.

Conclusion: $f(x)$ is continuous on $[3,\infty )$.

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Key Formulas or Methods Used

• Domain of square root function: $\sqrt{u}$ is defined only for $u\ge 0$ (in real analysis)
• Continuity requirement: A function must be defined at every point in an interval to be continuous on that interval
• Composition of continuous functions: If $g$ is continuous at $x$ and $h$ is continuous at $g(x)$, then $h\circ g$ is continuous at $x$

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Summary of Steps

1. Determine the domain: Solve $x^2-9\ge 0$ to find where $f(x)$ is defined, yielding $x\le -3$ or $x\ge 3$.
2. Analyze interval $[-3,3]$: Identify that $f(x)$ is undefined (imaginary) for $x\in (-3,3)$ because $x^2-9<0$, making the function discontinuous on this interval.
3. Analyze interval $[3,\infty )$: Verify that $x^2-9\ge 0$ for all $x\ge 3$, so $f(x)$ is defined everywhere on this interval.
4. Confirm continuity: Recognize $f(x)$ as a composition of continuous functions, ensuring continuity on $[3,\infty )$.