Let f ( x ) be strictly increasing and strictly convex, and g ( x ) be strictly increasing and concave, then is also true that h ( x ) = f ( x ) g ( x ) is convex?

Question

Let   f  (  x  ) be strictly increasing and strictly convex, and   g  (  x  ) be strictly increasing and concave, then is also true that   h  (  x  )  =  f  (  x  )  g  (  x  ) is convex?

Savanna Smith · Accepted Answer

For each product fg  f  g, that   (  f  g      )    ′    =      f    ′    g  +  f      g    ′   and   (  f  g      )    ″    =      f    ″    g  +      f    ′        g    ′    +      f    ′        g    ′    +  f      g    ″  .That means   (  f  g      )    ″    =      f    ″    g  +  2      f    ′        g    ′    +  f      g    ″  . We know   2      f    ′        g    ′    &amp;gt;  0 and       f    ″    &amp;gt;  0,       g    ″    ≥  0.If, as you say in your comment,   f and   g are also strictly positive, then       f    ″    g  &amp;gt;  0 and   f      g    ″    ≥  0, so   (  f  g      )    ″    =      f    ″    g  +  2      f    ′        g    ′    +  f      g    ″    &amp;gt;  0 and we proved   h is convex.

Cael Dickerson · Accepted Answer

f  (  x  )  =      x    2    and   g  (  x  )  =  −      x          −      2       over the domain   x  ∈            R        +  .If you insist on   g  &amp;gt;  0, we can restrict the target domain to   x  ∈  (  0  ,  1  ) and get   f  (  x  )  =      x    2    +  3      /    2  +  ϵ and   g  (  x  )  =  −      x    2    +  2  x, which has   h  (  x  ) become concave down everywhere. without the   ϵ,   h  (  x  )  =  −      x    4    +  2      x    3    −      3    2        x    2    +  3  x is actually straight at   x  =  1      /    2.

Let f(x) be strictly increasing and strictly convex, and g(x) be strictly increasing and concave, then is also true that h(x)=f(x)g(x) is convex?

Answered question

Answer & Explanation

New Questions in Calculus 1