Is there a quick method for sketching the general shape of $f(x)=x^2 e^x$ by inspection?

Question

I would like to know if there is a way to sketch the graph of $f(x)=x^2 e^x$ without finding the axis intercepts, asymptotes, and differentiating to find the maxima/minima etc.

I was informed that in trying to sketch $f(x)$ it is helpful to sketch the separate functions first.

So the graph of $e^x$ is

and the graph of $x^2$ is

the graph of $f(x)=x^2 e^x$ is

How could I sketch this graph myself just by inspection (without using a computer or scientific calculator)?

Are you asking how to intuitively know what this looks like without doing any calculation? — Ben Longo, Nov 27 '15 at 05:24
mentioned this in my answer but once you try doing something in mathematical/matlab/maple it would be worth putting this question on the respective SE or elsewhere. — djechlin, Nov 27 '15 at 05:30
@djechlin Mathematica counter-part has been deleted, I never used that site before and had no idea that they specialize in computer software. — BLAZE, Nov 27 '15 at 05:53
@djechlin This question is completely unrelated to the use of Mathematica. — Ben Longo, Nov 27 '15 at 05:54

score 1 · Answer 1 · answered Nov 27 '15 at 05:26

I think there's good reason to believe the answer is "no."

The product rule itself is fairly complicated; you can barely tell the derivative "at a glance" by looking at the algebra and I'm not aware of any more compact geometric interpretation.
You'd be able to infer a parabola $x^2$ from a line $x$, which sounds absurd.

Not sure how someone might trump my "no" vote but the answer will have to work within those constraints. I'm not optimistic.

Taking it off the graph you could do something weirder like use a 3D graph where the $y$-axis is $f(y)$ and the $x$-axis is $g(x)$ and $z = xy$. Obviously the scales will hide the graphs in this case, but sometimes this is useful; i.e. that's exactly what a log-scale is. Consider using color. (Also try this on the Mathematica SE?)

score 1 · Accepted Answer · answered Nov 27 '15 at 05:37

1

I think it's possible to get a pretty good idea of what this looks like intuitively. First off we know that $x^2e^x\ge 0$ and that it hits the point $(0,0)$. We also know that $e^x$ dominates $x^2$, so as $x$ becomes large, the shape of the curve will pretty much look like $e^x$, but for smaller values of $x$, the function will grow a bit faster. The horizontal asymptote at $y=0$ of $e^x$ will also remain because of this. In order to form a rough sketch, I would plot a few points when $x=0, 1, 2$ and then interpolate visually from there with a pretty tight horizontal asymptote at $y=0$ as $x\to-\infty$.

We can confirm this intuition by looking at a plot of $e^x$ compared to $x^2e^x$.

For larger $x$ we see the very similar shape.

answered Nov 27 '15 at 05:37

Ben Longo

1,068

Not to be cynical, but there's nothing about this that is 1) more general than handling $x^2e^x$ that doesn't 2) handle any way to combine two functions. It gives no intuition on products. – djechlin Nov 27 '15 at 05:49
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I don't think the question is about products of functions in general and I think it is impossible to provide a 'general intuition' for visualizing function products because the possibilities are literally endless. – Ben Longo Nov 27 '15 at 05:53
@Ben This relevance of this question will become clear very soon; as it stands this was simply a paradigm for what's coming next. – BLAZE Nov 27 '15 at 05:59

Is there a quick method for sketching the general shape of $f(x)=x^2 e^x$ by inspection?

2 Answers2