Can we say that ∀x∈R,g(x)&gt;0,g(x)&gt;0 or that (∃x∈R)g(x)=0.

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Can we say that ∀x∈R,g(x)&amp;gt;0,g(x)&amp;gt;0 or that (∃x∈R)g(x)=0.

PietAppeteoth22m · Accepted Answer

Using the &quot;vertex form&quot; of the polynomials, we are takingf(x)=ax2+bx+c=a⋅(x+b2a)2+(c−b24a)&amp;gt;0,which then requires c−b24a&amp;gt;0. [Naturally, this is equivalent to the discriminant condition.]For  g(x) = f(x)+f′(x)+f″(x) = ax2+(2a+b)x+(2a+b+c)  ,then g(x)  =  a·(x + (2a + b)2a)2 + ( [2a+b+c] − (2a + b)24a)=  a·(x + [ b2a+1 ])2 + ( [2a+b+c] − [a+b+b24a] )=  a·(x + [ b2a+1 ])2 + ( a+[ c−b24a ] )  .The vertex of the parabola corresponding to  g(x)  is therefore always one unit &quot;to the left&quot; and a units &quot;above&quot; the vertex for f(x) with a&amp;gt;0, guaranteeing that g(x)&amp;gt;0.

produsint8l5 · Accepted Answer

We can conclude g(x)&amp;gt;0 everywhere. If f(x)&amp;gt;0 then a&amp;gt;0, and since the coefficient of x2 in g(x) is also a, and the discriminant of g is negative, we must have g(x)&amp;gt;0 for all x.

Can we say that \(\displaystyle\forall{x}\in{\mathbb{{{R}}}},{g{{\left({x}\right)}}}{>}{0},{g{{\left({x}\right)}}}{>}{0}\) or that

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