How do you determine the values of x at which #sqrt(x^2 + 9)# is differentiable?

To determine the values of ( x ) at which ( \sqrt{x^2 + 9} ) is differentiable, we need to consider the domain of the function ( \sqrt{x^2 + 9} ). Since the square root function is defined only for non-negative values, we must ensure that the expression inside the square root, ( x^2 + 9 ), is non-negative.

For ( x^2 + 9 ) to be non-negative, ( x^2 ) must be greater than or equal to -9. However, since ( x^2 ) is always non-negative for real numbers, the expression ( x^2 + 9 ) will always be greater than or equal to 9. Thus, the function ( \sqrt{x^2 + 9} ) is defined for all real numbers ( x ).

Since the function is defined for all real numbers, it is differentiable for all values of ( x ) in its domain, which is the set of real numbers. Therefore, ( \sqrt{x^2 + 9} ) is differentiable for all real values of ( x ).