Question

let ( f ) be the function given by ( f(x) = \frac{x^2 - 9}{sin x} ) on the closed interval (0, 5). of the following intervals, on which can the mean value theorem be applied to ( f )?

i. (1, 3), because ( f ) is continuous on (1, 3) and differentiable on ((1, 3)).

ii. (4, 5), because ( f ) is continuous on (4, 5) and differentiable on ((4, 5)).

iii. (1, 4), because ( f ) is continuous on (1, 4) and differentiable on ((1, 4)).

a none
b i only
c i and ii only
d i, ii, and iii

Explanation:

Step1: Recall Mean Value Theorem (MVT) conditions

MVT requires a function \( f(x) \) to be continuous on \([a,b]\) and differentiable on \((a,b)\). The function here is \( f(x)=\frac{x^2 - 9}{\sin x}=\frac{(x - 3)(x + 3)}{\sin x} \). The denominator \(\sin x = 0\) when \( x = n\pi, n\in\mathbb{Z}\). In \([0,5]\), \(\sin x=0\) at \( x = 0,\pi\approx3.14, 2\pi\approx6.28\) (but \( 2\pi>5\)). So critical points are \( x = 0,\pi\).

Step2: Analyze Interval I: \([1,3]\)

• Continuity: On \([1,3]\), \(\sin x

eq0\) (since \(\pi\approx3.14\), so in \([1,3]\), \( x
eq\pi\) as \( 3<\pi\approx3.14\)? Wait, \( 3\approx3 < \pi\approx3.14\), so \(\sin x\) is non - zero on \([1,3]\). The numerator \( x^2 - 9\) is a polynomial (continuous everywhere), and the denominator \(\sin x\) is continuous and non - zero on \([1,3]\). So \( f(x)\) is continuous on \([1,3]\) (quotient of continuous functions with non - zero denominator).

• Differentiability: The derivative of \( f(x)\) can be found using the quotient rule \( f^\prime(x)=\frac{(2x)\sin x-(x^2 - 9)\cos x}{\sin^2 x}\). On \((1,3)\), \(\sin x

eq0\), so the denominator \(\sin^2 x
eq0\), and the numerator is a combination of differentiable functions (polynomial, sine, cosine), so \( f(x)\) is differentiable on \((1,3)\). So MVT can be applied to I.

Step3: Analyze Interval II: \([4,5]\)

• Continuity: On \([4,5]\), \(\sin x

eq0\) (since \(\pi\approx3.14\), \( 2\pi\approx6.28>5\), so in \([4,5]\), \( x\) is not a multiple of \(\pi\)). The numerator \( x^2 - 9\) is continuous, denominator \(\sin x\) is continuous and non - zero on \([4,5]\). So \( f(x)\) is continuous on \([4,5]\).

• Differentiability: Using the quotient rule, the derivative \( f^\prime(x)=\frac{(2x)\sin x-(x^2 - 9)\cos x}{\sin^2 x}\). On \((4,5)\), \(\sin x

eq0\), so the denominator is non - zero, and the numerator is differentiable. So \( f(x)\) is differentiable on \((4,5)\). So MVT can be applied to II.

Step4: Analyze Interval III: \([1,4]\)

• Continuity: In the interval \([1,4]\), \( x = \pi\approx3.14\in[1,4]\). At \( x=\pi\), \(\sin x = 0\), so \( f(x)\) is not defined at \( x = \pi\) (since denominator is zero). Therefore, \( f(x)\) is not continuous on \([1,4]\) (because it has a discontinuity at \( x=\pi\)). So MVT cannot be applied to III.

Answer:

C. I and II only