Question

question
consider the function (f(x)=\frac{3}{sqrt{25 - x^{2}}}) over the interval (2,4). does the extreme - value theorem guarantee the existence of an absolute maximum and minimum for (f(x)) on this interval?
select the correct answer below:
yes
no
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Explanation:

Step1: Check conditions for Extreme - Value Theorem

The Extreme - Value Theorem states that if a function \(y = f(x)\) is continuous on a closed interval \([a,b]\), then \(f(x)\) has both an absolute maximum and an absolute minimum on \([a,b]\). The function \(f(x)=\frac{3}{\sqrt{25 - x^{2}}}\) is continuous on the interval \([2,4]\) since the denominator \(\sqrt{25 - x^{2}}
eq0\) for \(x\in[2,4]\) (when \(x = 2\), \(\sqrt{25 - 4}=\sqrt{21}
eq0\); when \(x = 4\), \(\sqrt{25 - 16}=3
eq0\)). So the Extreme - Value Theorem guarantees the existence of an absolute maximum and minimum. The answer is Yes.

Answer:

Yes