Recent content by Shai

At the end it wasn't as hard as I thought, thank you very much and sorry for taking that long to understand.

So because $$f ''(x) $$ is a quadratic polynomial the x must have two solutions (greater than 0) in order to have inflection points?

Yes but why the sign doesn't change if the square root is equal to 0 but works with any other result of the square root? $$x=\frac{1}{2}(±√(a2−4b+8)−a−4)$$ $$x=\frac{1}{2}(±√(0)-a-4)$$

A point where the concavity of a continuous function change

Sorry but I don't understand why It works if the square root is 0, what's left in the x doesn't matter?: $$\frac{-4-a}{2}$$ The whole x shouldn't be equal to 0? Yes

The original function: $$f(x) = (x^2+ax+b)(e^x)$$ Then: $$f ' (x)= (x^2 +ax + b)(e^x)+(e^x)(2x+a)$$ $$f ' (x)= e^x(x^2+2x+ax+a+b)$$ Then : $$f '' (x)= (x^2+2x+ax+a+b)(e^x) + e^x(2x+2+a)$$ $$f '' (x)= e^x(x^2+4x+ax+2a+b+2)$$ And isolated the x with the formula for quadratic polynomial...

I still don't understand what do you mean by two zeros? The 0 of the square root and the f''=0 ? I am not able to visualize that in my head

<Moderator's note: Moved from a technical forum and thus no template.> The title isn't complete this is what I meant to say: Determine the values of aa and bb where the function has inflection points (x2+ax+b)(ex) I made the second derivative $$f''(x) = 2 e^x + 2 a e^x + b e^x + 4 e^x x + a...