Equality of a quadratic functionLet f : R → R an arbitrary function and g : R → R a quadratic function with the following property:For any m and n the equation f ( x ) = m x + n has a solution iff the equation g ( x ) = m x + n has a solution.Prove that f and g are equal.

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Equality of a quadratic functionLet   f  :      R    →      R   an arbitrary function and   g  :      R    →      R   a quadratic function with the following property:For any m and n the equation   f  (  x  )  =  m  x  +  n has a solution iff the equation   g  (  x  )  =  m  x  +  n has a solution.Prove that f and g are equal.

embutiridsl · Accepted Answer

Step 1WLOG let g curves up.1.   f  (      x    0    )  ≮  g  (      x    0    ). Suppose   f  (      x    0    )  &amp;lt;  g  (      x    0    ). Call tangent line of g at       x    0   as h. Then   h  −  ϵ does not meet g at any point. (Because g is quadratic) However, for a certain   ϵ, this line pass through   f  (      x    0    ). Contradiction. Since the choice of       x    0   is arbitrary,   f  ≥  g.Step 22.   f  (      x    0    )  ≯  g  (      x    0    ). From (1),   f  (  x  )  ≥  g  (  x  )  &amp;gt;  h  (  x  )     ∀  x  ≠      x    0  . Suppose   f  (      x    0    )  &amp;gt;  g  (      x    0    ), h does not meet any point on f, however h meet g at       x    0  . Contradiction.Therefore   f  (      x    0    )  =  g  (      x    0    ). Since the choice of       x    0   is arbitrary,   f  =  g.

Jefferson Booth · Accepted Answer

Step 1The following solution is a more explicit (algebraic) description of the geometric ideas from the previous solution.Suppose   g  (  x  )  =  a      x    2    +  b  x  +  c, and suppose that   f  ≠  g. Assume that   a  &amp;gt;  0 (otherwise you can consider -g and -f). Consider two cases:Case 1:   f  (  x  )  ≥  g  (  x  ) for all   x  ∈      R  . Since   f  ≠  g, there exists       x    0    ∈      R   such that   f  (      x    0    )  &amp;gt;  g  (      x    0    ). Let   m  =      g    ′    (      x    0    )  =  2  a      x    0    +  b, and   n  =  g  (      x    0    )  −  m      x    0    =  c  −  a      x    0    2  . Then the equation   g  (  x  )  =  m  x  +  n has a solution at       x    0  . However,   f  (      x    0    )  &amp;gt;  g  (      x    0    )  =  m      x    0    +  n, and   f  (  x  )  −  m  x  −  n  ≥  g  (  x  )  −  m  x  −  n  =  a  (  x  −      x    0        )    2    &amp;gt;  0 for   x  ≠      x    0  . Therefore the equation   f  (  x  )  =  m  x  +  n has no solutions in R.Case 2. There exists       x    0    ∈      R   such that   g  (      x    0    )  &amp;gt;  f  (      x    0    ). In this case, let   m  =      g    ′    (      x    0    )  =  2  a      x    0    +  b and let   n  =  f  (      x    0    )  −  m      x    0    =  f  (      x    0    )  −  2  a      x    0    2    −  b      x    0  . Then the equation   f  (  x  )  =  m  x  +  n has a solution at       x    0  , but   g  (  x  )  −  m  x  −  n  =  a  (  x  −      x    0        )    2    +  (  g  (      x    0    )  −  f  (      x    0    )  ), which is never zero (since   a  &amp;gt;  0 and   g  (      x    0    )  &amp;gt;  f  (      x    0    )), so the equation   g  (  x  )  =  m  x  +  n has no solutions in R.Step 2Hence we have shown that if   f  ≠  g, then there exist m,n such that one of the equations   f  (  x  )  =  m  x  +  n or   g  (  x  )  =  m  x  +  n has a solution and the other does not.

Equality of a quadratic function. Let f:R rightarrow R an arbitrary function and g:R rightarrow R a quadratic function with the following property:

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