Question

Let \(f\left( x \right) = 1 - {x^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}}\), show that \(f\left( { - 1} \right) = f\left( 1 \right)\) but there is no number \(c\) in \(\left( { - 1,1} \right)\) such that \(f'\left( c \right) = 0\). Why does this not contradict Rolle’s Theorem?

Short answer

It is verified that the \(f\left( { - 1} \right) = f\left( 1 \right)\).

And there is no value of \(c\)for which \(f'\left( c \right) = 0\).

As \(f'\left( 0 \right)\) does not exist, so \(f'\left( c \right) = 0\) does not contradict Rolle’s Theorem.

Step-by-step solution

Rolle’s Theorem

If a function\(f\)satisfies the below hypothesis, then there is a number \(c\) in\(\left( {a,b} \right)\), such that \(f'\left( c \right) = 0\),

1. When\(f\)is continuous on \(\left( {a,b} \right)\).
2. When\(f\)is differentiableon \(\left( {a,b} \right)\).
3. And \(f\left( a \right) = f\left( b \right)\)

Determine \(f\left( { - 1} \right) = f\left( 1 \right)\)

Find \(f\left( { - 1} \right)\) by substituting \( - 1\) for \(x\) into \(f\left( x \right) = 1 - {x^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}}\).

\(\begin{array}{c}f\left( { - 1} \right) = 1 - {\left( { - 1} \right)^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}}\\ = 0\end{array}\)

Find \(f\left( 1 \right)\) by substituting 1 for \(x\) into \(f\left( x \right) = 1 - {x^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}}\).

\(\begin{array}{c}f\left( 1 \right) = 1 - {\left( 1 \right)^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}}\\ = 0\end{array}\)

Hence, \(f\left( { - 1} \right) = f\left( 1 \right)\).

Find differentiation

Find the derivative of given function.

\(\begin{array}{c}f'\left( x \right) = \frac{d}{{dx}}\left( {1 - {x^{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}}} \right)\\ = - \frac{2}{3}{x^{ - {1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}}\end{array}\)

Find \(c\)

Now determine \(c\) by using \(f'\left( c \right) = 0\).

\(\begin{array}{c} - \frac{2}{3}{c^{ - {1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}} = 0\\\frac{1}{{{c^{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}}}} = 0\end{array}\)

It can be observed that, there is no value of \(c\) for \(f'\left( c \right) = 0\).

Determine existence of \(f'\left( 0 \right)\)

Substitute 0 for \(x\)into \(f'\left( x \right)\) and solve,

\(\begin{array}{c}f'\left( 0 \right) = - \frac{2}{3}{0^{ - {1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}}\\ = - \frac{2}{3} \cdot \frac{1}{0}\end{array}\)

Which implies, \(f'\left( 0 \right)\) does not exist.

Hence, this does not contradict Rolle’s Theorem, because \(f'\left( 0 \right)\) does not exist, so the given function is not differentiable on \(\left( { - 1,1} \right)\).