Question

Let \(f(x) = 1 - {x^{\frac{2}{3}}}\). Show that \(f( - 1) = f(1)\) but there is no number \(c\) in \(( - 1,1)\) such that \({f^\prime }(c) = 0\). Why does this not contradict Rolle's Theorem?

Short answer

The answer is stated below.

Step-by-step solution

Given

The function is \(f(x) = 1 - {x^{\frac{2}{3}}}\).

Concept of Rolle’s Theorem

Rolle's Theorem:

“Let\(f\)be a function that satisfies the following hypothesis:

1)\(f\)is continuous on the closed interval\(\left( {a,b} \right)\).

2)\(f\)is differentiable on the open interval\((a,b)\).

3)\(f(a) = f(b)\).

Then, there is a number\(c\)in\((a,b)\)such that,\(f'(c) = 0\).”

Check the third condition of Rolle’s Theorem

Find the values of\(f(x)\)for the end points of the given interval.

Substitute, \(x = - 1 \to f(x)\).

\(\begin{aligned}{c}f( - 1) &= 1 - {( - 1)^{\frac{2}{3}}}\\ &= 1 - 1\\ &= 0\end{aligned}\)

Substitute,\(x = 1 \to f(x)\).

\(\begin{aligned}{c}f(1) &= 1 - {(1)^{\frac{2}{3}}}\\ &= 1 - 1\\ &= 0\end{aligned}\)

Hence, it is proved that \(f( - 1) = f(1)\).

Check the second condition of Rolle’s Theorem

Find the derivative of\(f(x)\).

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {1 - {x^{\frac{2}{3}}}} \right)\\ &= 0 - \left( {\frac{2}{3}{x^{\frac{{ - 1}}{3}}}} \right)\\ &= - \frac{2}{3}\left( {\frac{1}{{{x^{\frac{1}{3}}}}}} \right)\end{aligned}\)

Thus,\({f^\prime }(c) = - \frac{2}{3}\left( {\frac{1}{{{c^{\frac{1}{3}}}}}} \right)\).

Notice that\({f^\prime }(c) = 0\)has no solution.

\(\begin{aligned}{c} - \frac{2}{3}\left( {\frac{1}{{{c^{\frac{1}{3}}}}}} \right) = 0\\\frac{1}{{{c^{\frac{1}{3}}}}} = 0\end{aligned}\)

Therefore, there is no existence of a point\(c\).

The function\(f(x)\)fails to satisfy the Rolle's Theorem conditions as the function\(f(x)\)is not differentiable in the open interval\(( - 1,1)\).

Logically, if the hypotheses (conditions) are false and the conclusion is either true or false, then the statement is true.

Therefore, it does not contradict the Rolle's Theorem.