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For what values of $a$ does the function $f(x)=e^x+ax^3$ have an inflection point?

It is an old question in my mind and wanted to bring it out.

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Try to differentiate $f$ twice. – Siminore May 8 '12 at 15:05
For which $a$'s does $f''$ change sign? – AD. May 8 '12 at 15:12
Nice question. I saw it right now. – Nancy Rutkowskie Jan 25 '13 at 9:50
up vote 3 down vote accepted

$$f''x=e^x+6ax$$ now,

$f''x=0$ when $e^x=-6ax$ ------[eqn1]

fram the graphs of $y=e^x$ and $y=-6ax$ it can be said that a solution to this will surely exist if $a>0$ ,

but when $a<0$ we things get complicated,

if a is too small the line $y=-6ax$ may not meet $e^x$ at all,

if a is too large, the line will meet the curve twice,

so we need to find the borderline case, when the line just touched the curve, i.e it is tangent to $e^x$

Let the line touch the curve at some $x=x_1$

slope of the line $y=-6ax$ is $-6a$

slope of $y=e^x$ at for some $x=x_1$ is $e^{x_1}$

so we have $-6a=e^{x_1}$

also since the line and the curve are touching we have,


from this and previous results we have,


this gives the solution either $a=0$ or $x=1$

putting x=1 in [eqn1] we get,

$e=-6a$ or $a =\frac{-e}{6}$

hence the complete and final solution is ,

$$a \in [0,\infty)\cup(-\infty ,\frac{-e}{6}] $$

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$$ f(x) = e^x + ax^3. $$ $$ f'(x) = e^x + 3ax^2. $$ $$ f''(x) = e^x + 6ax. $$

Suppose first that $a>0$.

Since $f''(x)\to\infty$ as $x\to\infty$ and $f''(x)\to-\infty$ as $x\to-\infty$, and $f''$ is everywhere continuous, there must be a point where $f''$ changes signs. And since $f'''(x)= e^x+6a$ is everywhere positive, $f''$ must be a strictly increasing function and hence can have only one point where it is equal to $0$, and there it must change signs.

Therefore there is only one solution to $e^x+6ax=0$, and that is where the inflection point is. I don't think there's a closed-form solution, but you could apply Newton's method to solve this equation.

If $a=0$, then $f''(x)= e^x$, and this is always positive, so there are no inflection points.

Now suppose $a<0$. Since $e^x\to0$ as $x\to-\infty$ and $6ax\to\infty$ as $x\to-\infty$, we have $f''(x)\to\infty$ as $x\to-\infty$. But $f''(x)\to \infty$ as $x\to\infty$ as well. Therefore the behavior of $f''(x)$ as $x\to\pm\infty$ does not necessitate any changes in sign of $f''(x)$. If $a$ is a long way below $0$, you get a steep line plunging downward faster than $e^x$ grows when $x$ is near $0$, and $f''$ will become negative at some point where $x>0$, and then will become positive again to the right of that point. You'll have two inflection points. I think Newton's method will find them. But if $a$ is negative and closer to $0$, the line going below $0$ will not move fast enough to outweigh the $e^x$ that is growing, and the graph of $f''$ will never go below the $x$-axis. How do we know which values of $a$ are far enough below $0$ that there will be two inflection points? So far all I can suggest is numerical experiments.

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I is so late for thanking you, but Thank you very much. +1 – Babak S. Jan 23 '14 at 16:27

Actually there is a very nice closed-form solution to this problem.

As shown $f^{\prime\prime}(x)=e^x+6ax$. And inflection points are then solution to equation $$e^x=-6ax.$$

This have an analytical solution using so-called Lambert W function, i.e., the inflection points are: $$x^\star = -W\left(\frac{1}{6a}\right)$$

From conditions on existence of Lambert function we know that if $\frac{1}{6a}\geq -\frac{1}{e}$, then $x\in\mathbb{R}$, otherwise $x\in\mathbb{C}$.

This leads to the same conditions on $a$ as Tomarinator mentioned, i.e., if $a\in(-\infty,-e/6]\cup(0,\infty)$, then $x\in\mathbb{R}$. For other $a$ (except $0$), $x\in\mathbb{C}$, where in both cases, $x^\star=-W\left(\frac{1}{6a}\right)$.

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