Question-51
Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be a twice-differentiable function and suppose its second derivative satisfies \(f^{\prime \prime }( x) >0\) for all \(x\in \mathbb{R}\). Which of the following statements is/are ALWAYS correct?
Since \(f^{\prime \prime }( x) >0\) for all \(x\in \mathbb{R}\), \(f^{\prime } :\mathbb{R}\rightarrow \mathbb{R}\) has to be a strictly increasing function. Since \(f^{\prime }\) is strictly increasing, it has to be one-one. That is, \(f^{\prime }( x) =f^{\prime }( y) \Longrightarrow x=y\). To put it in another way, there do not exist \(x,y\) with \(x\neq y\) such that \(f^{\prime }( x) =f^{\prime }( y)\). A specific case of this observation is option-(B).
Since \(f^{\prime }\) is strictly increasing, it can be equal to zero at most once. This means that \(f\) could have at most one local minimum. An example of this would be \(f:\mathbb{R}\rightarrow \mathbb{R}\), \(f( x) =x^{2}\).
Since \(f\) is differentiable and since the domain is \(\mathbb{R}\), a global minimum, if it exists, should also be a local minimum. Therefore, \(f\) has at most one global minimum. In fact, a stronger statement is this: for the function \(f\) given in this question, if \(x=a\) is a minimum, then \(x=a\) is both local and global.
Finally, \(f(x) = e^{x}\) is a counter-example for option-(A). It is a function that doesn’t have a minimum, either local or global.