I need help in this please !!! Consider the function f(x)=4−4x2/3 on the interval [−1,1]

Question

Consider the function f(x)=4−4x^2/3 on the interval [−1,1]

Which of the three hypotheses of Rolle's Theorem fails for this function on the inverval?

(a) f(x) is continuous on [−1,1].

(b) f(x) is differentiable on (−1,1).

(c) f(−1)=f(1).

Answer 1

Hi Mo! The correct answer to this question is (B). Let me explain why...

(A) is a true statement. The function f (x) = 4 - 4x ^ ( 2 / 3 ) is continous on the interval [-1, 1] because there are no discontinuities (there are no asymptotes, unbounded y values to ±∞, and such).

(B) is a false statement. Taking the derivative of the function leads to f ' (x) = ( - 8 / 3 ) * x ^ ( -1 / 3 ), or

f ' (x) = ( - 8 ) / ( 3 * x ^ ( 1 / 3 ) ). Notice that if x = 0, which - 1 < 0 < 1, then f ' (0) would be undefined because you cannot have a number -8 divided by 0. Because x = 0 is within ( -1, 1) domain, the function, at that domain, fails to be differentiable.

(C) is a true statement. 4 - 4 * ( -1 ) ^ ( 2 / 3 ) IS equal to 4 - 4 * (1) ^ ( 2 / 3 ).

Therefore, (B) is a FALSE statement!>>>>>>>> (ANSWER) Let me know if you want additional help!