Question

let ( f ) be the function given by ( f(x) = \frac{sin x cos x}{(x^2 - 4)} ) on the closed interval ( -2pi, 2pi ). on which of the following closed intervals is the function ( f ) guaranteed by the extreme value theorem to have an absolute maximum and an absolute minimum?
a ( -2pi, 2pi )
b ( -2, 2 )
c ( -1, 1 )
d ( left \frac{pi}{2}, pi
ight )

Explanation:

Step1: Recall Extreme Value Theorem

The Extreme Value Theorem states that if a function \( f \) is continuous on a closed interval \([a, b]\), then \( f \) attains an absolute maximum and an absolute minimum on \([a, b]\). So we need to check where \( f(x)=\frac{\sin x\cos x}{x^{2}-4} \) is continuous.

Step2: Find Discontinuities of \( f(x) \)

A rational function \( \frac{g(x)}{h(x)} \) is discontinuous where \( h(x) = 0 \). Here, \( h(x)=x^{2}-4=(x - 2)(x + 2) \), so \( h(x)=0 \) when \( x = 2 \) or \( x=-2 \). So \( f(x) \) is discontinuous at \( x = 2 \) and \( x=-2 \).

Step3: Analyze Each Interval

• Option A: \([-2\pi, 2\pi]\)

\( 2\pi\approx6.28 \), so \( x = 2 \) and \( x=-2 \) are inside \([-2\pi, 2\pi]\). Thus, \( f(x) \) is discontinuous on this interval, so Extreme Value Theorem does not apply.

• Option B: \([-2, 2]\)

The interval contains \( x = 2 \) and \( x=-2 \), where \( f(x) \) is discontinuous. So Extreme Value Theorem does not apply.

• Option C: \([-1, 1]\)

The interval \([-1, 1]\) does not contain \( x = 2 \) or \( x=-2 \) (since \( - 2<-1 \) and \( 1 < 2 \)). Also, \( \sin x\cos x=\frac{1}{2}\sin(2x) \) is continuous everywhere, and the denominator \( x^{2}-4
eq0 \) on \([-1, 1]\). So \( f(x) \) is continuous on \([-1, 1]\). By Extreme Value Theorem, \( f(x) \) has absolute max and min here.

• Option D: \([\frac{\pi}{2},\pi]\)

\( \pi\approx3.14>2 \), so \( x = 2 \) is inside \([\frac{\pi}{2},\pi]\) (since \( \frac{\pi}{2}\approx1.57<2<\pi\approx3.14 \)). Thus, \( f(x) \) is discontinuous on this interval, so Extreme Value Theorem does not apply.

Answer:

C. \([-1,1]\)