- #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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