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.

9. \(f\left( x \right) = 2{x^2} - 4x + 5\), \(\left( { - 1,3} \right)\).

Short answer

The given function satisfies Rolle’s Theorem. The value of \(c\) is \(1\).

Step-by-step solution

Rolle’s Theorem

Rolle’s Theorem states that if a function \(f\) satisfies the following:

1. Continuous on an interval \(\left( {a,b} \right)\).

2. Differentiable on \(\left( {a,b} \right)\).

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

Then there exists atleast one point \(c \in \left( {a,b} \right)\) such that \(f'\left( c \right) = 0\).

Verification of Rolle’s Theorem

The given function is \(f\left( x \right) = 2{x^2} - 4x + 5\) on \(\left( { - 1,3} \right)\).

Since all the polynomials are continuous on \(\mathbb{R}\) hence the given function is continuous on \(\left( { - 1,3} \right)\).

Again, since all the polynomials are differentiable on \(\mathbb{R}\), so the given function is differentiable on \(\left( { - 1,3} \right)\).

Also \(f\left( { - 1} \right) = f\left( 3 \right) = 11\)

Hence the given function satisfies Rolle’s Theorem.

Then there exist \(c \in \left( { - 1,3} \right)\), Such that \(f'\left( c \right) = 0\).

\(\begin{aligned}{c}f'\left( c \right) &= 0\\4c - 4 &= 0\\4c &= 4\\c &= 1\end{aligned}\)

Hence \(c = 1\) and \(c \in \left( { - 1,3} \right)\).