Question

Suppose that \(f\) and \(g\) are continuous on \(\left( {a,b} \right)\) and differentiable on \(\left( {a,b} \right)\). Suppose also \(f\left( a \right) = g\left( a \right)\) and \(f'\left( x \right) < g'\left( x \right)\) for \(a < x < b\). Prove that \(f\left( b \right) < g\left( b \right)\).

Short answer

It is proved that \(f\left( b \right) < g\left( b \right)\).

Step-by-step solution

The mean value theorem

Let \(f\) be a function that is continuous on the closed interval \(\left( {a,b} \right)\) and differentiable on the open interval \(\left( {a,b} \right)\). Then there is a number \(c \in \left( {a,b} \right)\) such that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( b \right)}}{{b - a}}\).

Proof the statement

Let \(h\left( x \right) = f\left( x \right) - g\left( x \right)\). It is given that \(f\left( a \right) = g\left( a \right)\). Therefore, \(h\left( a \right) = f\left( a \right) - g\left( a \right) = 0\).

Now, \(f\) and \(g\) are continuous on \(\left( {a,b} \right)\) and differentiable on \(\left( {a,b} \right)\). So, \(h\left( x \right)\) also satisfies the same conditions. So, by Mean Value Theorem, there is a number \(c\) in the interval \(\left( {a,b} \right)\) such that \(h'\left( c \right) = \frac{{h\left( b \right) - h\left( a \right)}}{{b - a}}\).

Solve the above equation as follows:

\(\begin{aligned}{c}h'\left( c \right) &= \frac{{h\left( b \right) - h\left( a \right)}}{{b - a}}\\h'\left( c \right) &= \frac{{h\left( b \right) - 0}}{{b - a}}\\h'\left( c \right)\left( {b - a} \right) &= h\left( b \right)\end{aligned}\)

Here, it is given that \(f'\left( x \right) < g'\left( x \right)\). So, \(h'\left( c \right) < 0\). That implies \(h'\left( c \right)\left( {b - a} \right) < 0 \Rightarrow h\left( b \right) < 0\).

So, \(f\left( b \right) - g\left( b \right) < 0\) implies \(f\left( b \right) < g\left( b \right)\).

Hence, proved that \(f\left( b \right) < g\left( b \right)\).