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rociomartinez9

30.09.2019 •

Mathematics

Explain why rolle's theorem does not apply to the function even though there exist a and b such that f(a) = f(b). (select all that apply.)
f(x) = sqrt((2-x(2/3))3) text(, [-1,1])

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Ответ:

maysahdabest

04.01.2022

Mathematics

has to be differentiable on [-1, 1] for the theorem to hold. This is not the case because the derivative doesn't exist at x=0

.

$f(x)=\sqrt{\left(2-x^{2/3}\right)^3}=\left(2-x^{2/3}\right)^{3/2}$
$\implies f'(x)=\dfrac32\left(2-x^{2/3}\right)^{1/2}\left(-\dfrac23x^{-1/3}\right)=-\dfrac{\left(2-x^{2/3}\right)^{1/2}}{x^{1/3}}$

but clearly f'(0)

is undefined.

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Ответ:

Nickanderson21

14.02.2020

Mathematics

Arc length = radius* angle theta (in radians)

So angle 90 degree = 90*pi/180 radians= pi/2 radians

So arc length = 7.5* pi/2 cm = 11.8cm

So option A is correct.

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Explain why rolle's theorem does not apply to the function even though there exist a and b such that f(a) = f(b). (select all that apply.) f(x) = sqrt((2-x**(2/3))**3) text(, [-1,1])

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Explain why rolle's theorem does not apply to the function even though there exist a and b such that f(a) = f(b). (select all that apply.)
f(x) = sqrt((2-x(2/3))3) text(, [-1,1])