Find the largest possible domain of the real valued function f(x)=dfrac√(4-x²)√(x²-9).

Exercise 1.3 · Q7

f(x)=\dfrac{\sqrt{4-x^2}}{\sqrt{x^2-9}}

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Step 1. For the numerator $\sqrt{4-x^2}$ to be real, need $4-x^2\ge0\Rightarrow x^2\le4\Rightarrow x\in[-2,2]$ .

Step 2. For the denominator $\sqrt{x^2-9}$ to be real AND nonzero (it can't be zero, since it's a denominator), need $x^2-9>0\Rightarrow x^2>9\Rightarrow x\in(-\infty,-3)\cup(3,\infty)$ .

Step 3. The domain of $f$ is the intersection of both requirements: $[-2,2]\cap\big((-\infty,-3)\cup(3,\infty)\big)$ . Since $[-2,2]$ lies entirely between $-3$ and $3$ , it shares no point with $x<-3$ or $x>3$ . The intersection is empty. …

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Q.Find the largest possible domain of the real valued function $f(x)=\dfrac{\sqrt{4-x^2}}{\sqrt{x^2-9}}$ .

Concept understanding — Intervals and Neighbourhoods

Q.Find the largest possible domain of the real valued function f(x)=4−x2x2−9f(x)=\dfrac{\sqrt{4-x^2}}{\sqrt{x^2-9}}f(x)=x2−9​4−x2​​.

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Q.Find the largest possible domain of the real valued function $f(x)=\dfrac{\sqrt{4-x^2}}{\sqrt{x^2-9}}$ .

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