# If $f(x)=e^{-x}x^n$, find the values of $n$ such that $f(x)$ has $0,1, 2$ or $3$ inflection points.

If $$f(x)=e^{-x}x^n$$, $$n\in \mathbb{Z}$$, find the values of $$n$$ such that $$f(x)$$ has $$0,1, 2$$ or $$3$$ inflection points. (That is, give the set of $$n$$ for each possible number of inflection points (0,1,2,3).)

This is a question that I found in a past exam paper for a math subject in Australia.

The previous parts of the question ask to find $$f''(x)$$ and to find $$x: f''(x)=0, x\neq 0$$

The answers are $$f''(x)= e^{-x} x^{n - 2} (n^2 - n (2 x + 1) + x^2)$$, and for the next part, $$x=n\pm \sqrt{n}$$ ($$\neq0$$).

So for the question I've asked, I actually did solve it, but in a fairly long winded way. As an inflection point is defined if the following conditions are met: $$f''(x)=0, \frac{f''(x+a)}{f''(x-a)}<0$$, (the last condition saying that the value of $$f''(x)$$ in the neighbourhood of the x-intercept of $$f''(x)$$ are opposite signs immediately to the left and right of it.), I did the following:

For each of the factors in $$f''(x)$$, consider when they would satisfy the alternating sign condition, then I essentially intersected each set of n-values to arrive at the answer.

0 POIs: $$n\in \mathbb{Z}^-\cup \{0\}$$

1 POI: $$n=1$$

2 POIs: $$n=2k, k\in \mathbb{Z}^+$$

3 POIs: $$n=2k+1, k\in \mathbb{Z}^+$$

The problem though is that this took too long, and the question was only worth 2 marks which really means that it shouldn't take more than a few minutes to figure out.

My question: is there are nicer less complex way of solving this? My method involved intersecting a bunch of sets and considering certain 'edge' cases, which doesn't seem very efficient at all.

• Note that if the first part specifically asked for no-zero roots, then the answer $n\pm\sqrt n$ sometimes (namely, for $n=0,1$) wrongly includes the to be excluded $x=0$. – Hagen von Eitzen Apr 14 at 10:19

Let's start from where you set the second derivative equal to zero: $$e^xx^{n-2}(x^2 - 2nx + (n^2 - n)) = 0$$. Notice that we can divide both sides by the $$e^x$$ because it's never zero, so $$x^{n-2}(x^2 - 2nx + (n^2 - n)) = 0$$, and now our equation is polynomial.

Now we can use one of the properties of the roots of a polynomial, which is that if we have a root with an even multiplicity, then the polynomial will have the same sign on either side, but if we have a root with an odd multiplicity, then the polynomial will have opposite signs on either side. So, we want to look for roots with odd multiplicities.

For the first term, $$x^{n-2}$$, if $$n > 2$$ we will have a root at $$x = 0$$. If $$n$$ is even, this will have even multiplicity, otherwise it will have odd multiplicity.

Let's look at the first term: $$x^{n-2} = 0$$. When $$n \leq 2$$, this trivially has no solutions. Otherwise, we get a root at $$x = 0$$ with even multiplicity if $$n$$ is even, otherwise odd multiplicity if $$n$$ is odd.

Now let's look at the second term: $$x^2 - 2nx + (n^2 - n) = 0$$. Let's look at the discriminant, which is the expression under the square root in the quadratic formula: $$(-2n)^2 - 4(1)(n^2 - n) = n$$. So when $$n < 0$$, we get no real roots, when $$n = 0$$ we get one double root, and when $$n > 0$$ we get two distinct real roots with multiplicity $$1$$.

There is a potential concern here with whether the roots from the two terms will match up, so let's consider if $$x = 0$$ is ever a root of the second term. If $$x = 0$$ is a root of the second term, then we must have $$n^2 - n = 0$$, so $$n$$ is $$0$$ or $$1$$. But when $$n$$ is $$0$$ or $$1$$, our expression is undefined at $$x = 0$$ due to division by zero, so we have to disregard this root.

So, for $$n \leq 0$$ we have no roots, and therefore no points of inflection.

When $$n = 1$$, the first term has no roots and the second term gives two roots with multiplicity $$1$$, but we have to ignore the root at $$x = 0$$ so we get one point of inflectin.

For $$n \geq 2$$, the second term gives us $$2$$ points of inflection for every value of $$n$$, and the first gives one additional point of inflection if $$n$$ is odd. This matches your answer.

• The expression under the square root in the quadratic formula is called the "discriminant" , no "determinant". – jjagmath Apr 14 at 10:44
• I did say determinant didn't I. My bad, I'll fix that now – Stephen Donovan Apr 14 at 10:45