Question

Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.

8) There exists a function f such that \(f\left( 1 \right) = - 2\), \(f\left( 3 \right) = 0\) and \(f'\left( x \right) > 1\) for all x.

Short answer

The given statement is False.

Step-by-step solution

Given

Suppose there is a function\(f\left( x \right)\)such that

\(f\left( 1 \right) = - 2\), \(f\left( 3 \right) = 0\)and \(f'\left( x \right) > 1\;\forall x\)

Consider the interval \(\left( {1,3} \right)\).

Since \(f'\left( x \right) > 1\;\forall x\).

\(\therefore \)f is differentiable.

\(\therefore \)f is continuous also.

 Apply mean value theorem

\(\therefore \)by mean value theorem

\(\exists c \in \left( { - 1,3} \right)\)

\(f'\left( c \right) = \frac{{f\left( 3 \right) - f\left( 1 \right)}}{2}\)

\(f'\left( c \right) = \frac{{0 - \left( { - 2} \right)}}{2}\)

\(f'\left( c \right) = 1......(1)\)

Since \(f'\left( x \right) > 1\;\forall x\)

\(\therefore f'\left( x \right) > 1\;\forall x\)

\(1 > 1\;\forall x\) (using 1)

\(\therefore \)this is a contradiction.

Hence this kind of function can never exist.

Hence our question statement is false.