Consider the functions

I. e^-x

II. x² – sin x

III. \(\sqrt {{x^3} + 1} \)

Which of the above functions is/are increasing everywhere in [0, 1]?

Concept:

A function f(x) is said to be increasing in the given interval if it’s first order differential \(f'\left( x \right) \ge 0\) holds for every point in the given interval.

Calculation:

Function I:

\(f\left( x \right) = {e^{ - x}}\therefore f'\left( x \right) = \; - {e^{ - x}}\)

\(\because {f'}\left( 0 \right) = - 1 < 0\)

Therefore, the function is non increasing.        

Function II:

\(f\left( x \right) = \;{x^2} - \sin x,\;f'\left( x \right) = 2x - \cos x\)

\(f'\left( 0 \right) = \;0 - \cos 0 = - 1 < 0\)

Therefore, this function is also non increasing.

Furthermore, since cosine function is periodic function, it cannot be strictly increasing in the given range

Function III:

\(f\left( x \right) = \sqrt {{x^3} + 1} \)

\({\rm{f'}}\left( {\rm{x}} \right) = \frac{{3{x^2}}}{{2\sqrt {{x^3} + 1} }}\)

\(f'\left( 0 \right) = 0,\;f'\left( 1 \right) > 0\)

Therefore, the function is increasing in the given interval.

Therefore \(f\left( x \right) = \sqrt {{x^3} + 1} \) function is the only increasing everywhere in [0, 1]

Hence Option(1) is the correct answer.