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For what values of a, b and c does the curve $y = ax^3 + 3x^2 + bx + cx + e^x$ have exactly one point of inflection? Two points of inflection? No points of inflection? Provide a numerical approximation for the lowest value of a for which there is no inflection point.

I'm really stumped on this question. I've tried finding the second derivative but couldn't think of anything else to do from that point. Any help would be greatly appreciated. Thanks!

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Points of inflection occur when the second derivative is zero. What is the second derivative of your curve? How can you assure there's only one zero such that we have $+/-$ on either side of the zero. (You're allowed other zeroes that don't change the inflection, i.e. $x=0$ of $y=x^2$). – Ian Coley Dec 10 '13 at 19:46
I've found the second derivative to be $y'' = 6ax + 6 + e^x = 0$ but I could not think of how to solve that. – DinoMint Dec 10 '13 at 19:48
What happens when $x=0,1,-1,-2, 2$ and so on? – wannadeleteacct Dec 10 '13 at 20:02
Given that $e^x$ is positive will drive you to a negative x. Since it won't be too small you can probably use an estimate for $e^x$ around that x. – half-integer fan Dec 10 '13 at 20:04
You don't need to solve $6ax+6+e^x=0$, @DinoMint, only figure out how many solutions it has. Think of it as intersecting the graph of $y=e^x$ with the line $y=-6ax-6$. If the slope $-6a$ is negative, there is exactly one intersection point, if it is positive you can find the tangent line to $y=e^x$ with the same slope ($-6a$) and see if this line, $y=-6ax-6$ is above or below the tangent. – Omar Antolín-Camarena Dec 10 '13 at 20:05
up vote 0 down vote accepted

As you stated in the comments, the second derivative is $$y'' = 6ax + 6 + e^x = 0$$

First, we note that if $a=0$ the equation becomes $$ 6 + e^x = 0$$ which has no real solutions, thus there are no inflection points.

Rewriting the equation as $$x = -\dfrac{6+e^x}{6a}$$ or $$x = -\dfrac1{a} \cdot \left(1 + \dfrac 16 e^x \right)$$ makes the behavior clearer.

If $a \gt 0$ we can see that $x$ must be negative and thus $e^x < 1$ and there will be a solution in the neighborhood of $x = -\dfrac1a$.

I will leave the case where $a \lt 0$, thus $x \gt 0$ for you to consider.

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thanks for the detailed help – DinoMint Dec 10 '13 at 21:26

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