- #1

Shai

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<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

$$f''(x) = 2 e^x + 2 a e^x + b e^x + 4 e^x x + a e^x x + e^x x^2$$

and isolate the x:

$$x = \frac{1}{2}\biggl(\pm\sqrt{a^2 - 4 b + 8} - a - 4\biggr)$$

And a guy told me this:

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Now, you have inflection points if $$√a2−4b+8$$ is real. Or $$a2−4b+8≥0$$.

However, If $$a2−4b+8=0$$ then f′′(x) does not change sign when x crosses the root, $$a2−4b+8>0$$.

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And now I am lost because I've trying to figure out why if the root is 0 then f''(x) doesn't change sign? What's left in the fraction doesn't matter?

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

$$f''(x) = 2 e^x + 2 a e^x + b e^x + 4 e^x x + a e^x x + e^x x^2$$

and isolate the x:

$$x = \frac{1}{2}\biggl(\pm\sqrt{a^2 - 4 b + 8} - a - 4\biggr)$$

And a guy told me this:

---------------------------------------

Now, you have inflection points if $$√a2−4b+8$$ is real. Or $$a2−4b+8≥0$$.

However, If $$a2−4b+8=0$$ then f′′(x) does not change sign when x crosses the root, $$a2−4b+8>0$$.

----------------------------------------

And now I am lost because I've trying to figure out why if the root is 0 then f''(x) doesn't change sign? What's left in the fraction doesn't matter?

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