Questions: Consider the function (f(x)=frac6sqrtx^2-16) over the interval ([3,5]). Does the extreme value theorem guarantee the existence of an absolute maximum and minimum for (f(x)) on this interval?

Solution Steps

To determine if the extreme value theorem guarantees the existence of an absolute maximum and minimum for the function \( f(x) = \frac{6}{\sqrt{x^2 - 16}} \) on the interval \([3, 5]\), we need to check if the function is continuous on the closed interval. The function is defined and continuous where the expression under the square root is positive, i.e., \( x^2 - 16 > 0 \). We will check the continuity of the function on the given interval.

The Extreme Value Theorem states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), then \( f(x) \) must attain both an absolute maximum and an absolute minimum on that interval.

The given function is \( f(x) = \frac{6}{\sqrt{x^2 - 16}} \). We need to determine if this function is continuous on the interval \([3, 5]\).

The function \( f(x) = \frac{6}{\sqrt{x^2 - 16}} \) is defined only when the expression under the square root is positive, i.e., \( x^2 - 16 > 0 \). Solving this inequality:

\[ x^2 - 16 > 0 \implies x^2 > 16 \implies x > 4 \quad \text{or} \quad x < -4 \]

Since we are considering the interval \([3, 5]\), we focus on \( x > 4 \). Therefore, the function is only defined for \( x \in (4, 5] \).

The function \( f(x) \) is not defined at \( x = 4 \), which is within the interval \([3, 5]\). Therefore, \( f(x) \) is not continuous on the entire interval \([3, 5]\).

Final Answer