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Math  /  Calculus

QuestionDetermine whether Rolle's theorem can be applied to ff on the closed interval [a,b][a, b]. (Select all that apply.) f(x)=x29−x2,[−10,10]f(x)=\frac{x^{2}}{9-x^{2}}, \quad[-10,10]
Yes, Rolle's theorem can be applied. No, because ff is not continuous on the closed interval [a,b][a, b]. No, because ff is not differentiable in the open interval (a,b)(a, b). No, because f(a)≠f(b)f(a) \neq f(b). If Rolle's theorem can be applied, find all values of cc in the open interval (a,b)(a, b) such that f′(c)=0f^{\prime}(c)=0. (Enter your answers as a comma-separated list. If Rolle's theorem cannot be applied, enter NA.) c=c=

Studdy Solution

STEP 1

1. We are given the function f(x)=x29−x2 f(x) = \frac{x^2}{9-x^2} .
2. The interval of interest is [−10,10][-10, 10].
3. We need to determine if Rolle's theorem applies, and if so, find the values of c c where f′(c)=0 f'(c) = 0 .

STEP 2

1. Check the continuity of f f on the closed interval [−10,10][-10, 10].
2. Check the differentiability of f f on the open interval (−10,10)(-10, 10).
3. Check if f(a)=f(b) f(a) = f(b) for a=−10 a = -10 and b=10 b = 10 .
4. If applicable, find the values of c c such that f′(c)=0 f'(c) = 0 .

STEP 3

Check the continuity of f(x)=x29−x2 f(x) = \frac{x^2}{9-x^2} on [−10,10][-10, 10].
The function f(x) f(x) is undefined where the denominator is zero. Solve 9−x2=0 9-x^2 = 0 to find the points of discontinuity.
9−x2=0 9 - x^2 = 0 x2=9 x^2 = 9 x=±3 x = \pm 3
Thus, f(x) f(x) is not continuous at x=3 x = 3 and x=−3 x = -3 .

STEP 4

Check the differentiability of f(x) f(x) on (−10,10)(-10, 10).
Since f(x) f(x) is not continuous at x=3 x = 3 and x=−3 x = -3 , it is also not differentiable at these points.

STEP 5

Check if f(a)=f(b) f(a) = f(b) for a=−10 a = -10 and b=10 b = 10 .
Calculate f(−10) f(-10) and f(10) f(10) :
f(−10)=(−10)29−(−10)2=1009−100=100−91 f(-10) = \frac{(-10)^2}{9 - (-10)^2} = \frac{100}{9 - 100} = \frac{100}{-91}
f(10)=1029−102=1009−100=100−91 f(10) = \frac{10^2}{9 - 10^2} = \frac{100}{9 - 100} = \frac{100}{-91}
Thus, f(−10)=f(10) f(-10) = f(10) .

STEP 6

Since f(x) f(x) is not continuous on [−10,10][-10, 10], Rolle's theorem cannot be applied.
The answer is:
No, because f f is not continuous on the closed interval [−10,10][-10, 10].
c=NA c = \text{NA}

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