Question

Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then find all numbers \(c\) that satisfy the conclusion of Rolle’s Theorem.

12. \(f\left( x \right) = x + {1 \mathord{\left/

{\vphantom {1 x}} \right.

\kern-\nulldelimiterspace} x},{\rm{ }}\left( {\frac{1}{2},2} \right)\)

Short answer

It is verified that the function \(f\left( x \right) = x + {1 \mathord{\left/

{\vphantom {1 x}} \right.

\kern-\nulldelimiterspace} x}\) satisfies all the three hypothesis of Rolle’s theorem and the value of \(c\) is \(1\).

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 \(f\left( a \right) = f\left( b \right)\)

Check for continuity

The given function \(f\left( x \right) = x + {1 \mathord{\left/

{\vphantom {1 x}} \right.

\kern-\nulldelimiterspace} x}\) is a rational function, and a rational function which is continuous on its domain \(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\), so it is also continuous on \(\left( {\frac{1}{2},2} \right)\).

Check for Differentiability

Find the derivative of given function.

\(\begin{aligned}{c}f'\left( x \right) &= \frac{d}{{dx}}\left( {x + {1 \mathord{\left/

{\vphantom {1 x}} \right.

\kern-\nulldelimiterspace} x}} \right)\\ &= 1 - \frac{1}{{{x^2}}}\\ &= \frac{{{x^2} - 1}}{{{x^2}}}\end{aligned}\)

The obtained function has the same domain \(\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\) and it is differentiable on it, so it is also differentiable on \(\left( {\frac{1}{2},2} \right)\).

Determine\(f\left( {\frac{1}{2}} \right) = f\left( 2 \right)\)

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

{\vphantom {1 x}} \right.

\kern-\nulldelimiterspace} x}\).

\(\begin{aligned}{c}f\left( {\frac{1}{2}} \right) &= \frac{1}{2} + {1 \mathord{\left/

{\vphantom {1 {\frac{1}{2}}}} \right.

\kern-\nulldelimiterspace} {\frac{1}{2}}}\\ &= \frac{1}{2} + 2\\ &= \frac{5}{2}\end{aligned}\)

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

{\vphantom {1 x}} \right.

\kern-\nulldelimiterspace} x}\).

\(\begin{aligned}{c}f\left( 2 \right) &= 2 + {1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}\\ &= \frac{5}{2}\end{aligned}\)

It can be observed that, \(f\left( {\frac{1}{2}} \right) = f\left( 2 \right)\).

Hence,the given function satisfies all three hypotheses of Rolle’s Theorem.

Find \(c\)

As the given function is differentiable, then find \(c\)by first solving \(f'\left( c \right) = 0\).

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

But \( - 1\) does not lie in the interval \(\left( {\frac{1}{2},2} \right)\). So \(c = 1\) is the only possible value.