Question

Let \(f\) and \(g\) represent differentiable functions such that \(f^{\prime \prime} \neq 0\) and \(g^{\prime \prime} \neq 0\). Prove that if \(f\) and \(g\) are positive, increasing, and concave upward on the interval \((a, b),\) then \(f g\) is also concave upward on \((a, b)\).

Short answer

The product of the functions \(f\) and \(g\), represented as \(fg\), is concave upwards on the interval \((a, b)\).

Step-by-step solution

Formulation of the Problem

We are given that \(f\), \(g\) and their second derivatives are positive on the interval \((a, b)\). The functions are also increasing and concave upward. We need to prove that the function \(fg\) is concave upward on the same interval.

Compute the Second Derivative of the Function \(fg\)

Using the product rule, we find that \((fg)'\) = \(f'g + fg'\). Then take the derivative of \((fg)'\) to find the second derivative \((fg)''\), which equals to \((f''g + 2f'g' + fg'')\).

Check the Concavity

A function is concave upward on an interval if its second derivative is positive over that interval. So, we need to check if \((fg)''\) is greater than 0 on the interval \((a, b)\). Since \(f\), \(g\), their first derivatives and second derivatives are positive on the interval by the given conditions, then it implies that \(f''g + 2f'g' + fg'' > 0\). Therefore, \(fg\) is concave upward on \((a, b)\).