If we study this function, we note that we can reduce the number of operations by introducing an intermediate variable
$$ a = x^2, $$leading to
$$ f(x) = f(a(x)) = b= \exp{a}. $$We now assume that all operations can be counted in terms of equal floating point operations. This means that in order to calculate \( f(x) \) we need first to square \( x \) and then compute the exponential. We have thus two floating point operations only.