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])

\(f(x)\) 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.

Answer Link

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])

Verified Answer :

Related Questions

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])