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])

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

[tex]f(x)=\sqrt{\left(2-x^{2/3}\right)^3}=\left(2-x^{2/3}\right)^{3/2}[/tex]
[tex]\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}}[/tex]

but clearly [tex]f'(0)[/tex] is undefined.

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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])