Let f : R + → R + and g : R + → R + be two strictly increasing, strictly concave and twice differentiable functions with f ( 0 ) = g ( 0 ) = 0 and f ′ ( 0 ) = g ′ ( 0 ) &gt; 0. We have f ( x ) ≤ g ( x ). Is it true that f ′ ( x ) ≤ g ′ ( x ) for any x in the domain?

Question

Let   f  :            R        +    →            R        +   and   g  :            R        +    →            R        +   be two strictly increasing, strictly concave and twice differentiable functions with   f  (  0  )  =  g  (  0  )  =  0 and       f    ′    (  0  )  =      g    ′    (  0  )  &amp;gt;  0. We have   f  (  x  )  ≤  g  (  x  ). Is it true that       f    ′    (  x  )  ≤      g    ′    (  x  ) for any   x in the domain?

Lena Gomez · Accepted Answer

No. Let   f  (  x  )  =  x for   0  ≤  x  &amp;lt;  1 and   f  (  x  )  =  1  +  x      /    4 for   x  &amp;gt;  1 (we can modify near   x  =  1 to make it smooth). Then g could be a similar piecewise linear function that starts out with a large slope on   [  0  ,  1      /    2  ] and has slope   1      /    2, say, on   (  1      /    2  ,  ∞  ).

Let f:R+->R^+ and g:R+->R^+ be two strictly increasing, strictly concave and twice differentiable functions with f(0)=g(0)=0 and f′(0)=g′(0)>0. We have f(x)<=g(x). Is it true that f′(x)<=g′(x) for any x in the domain?

Answered question

Answer & Explanation

New Questions in Calculus 1