Show that $f(x) = 2x \ln(x) - x^2$ has a point of inflection at $x=1$

Question

Show that $f(x) = 2x \ln(x) - x^2$ has a point of inflection at $x=1$.

$f'(x) \vert_{x=1} = \left[2 + 2 \ln(x) - 2x \right]\vert_{x=1} = 0$. Since $f'(x) < 0$ on either side of $x=1$, we can say that it can not be a local maxima or a local minima (as the derivatives would be of opposite signs on either side of $x=1$).

I am not sure if this concludes that $x=1$ is a point of inflection or I have to show something more. In particular, can a critical point be anything other a local maxima/minima and a point of inflection?

Answer 1

Yes you are right, we need to show that the second derivative is equal to zero (necessary condition):

$$f(x) = 2x \ln(x) - x^2\implies f''(x)=\frac2 x-2,\; f''(1)=0$$

and that the third derivative is not equal to zero:

$$f'''(x)=-\frac2 {x^2},\; f'''(1)=-2$$

therefore we have a (stationary) inflection point at $x=1$.

Refer also to:

• Do inflection points of $f(x)$ give $f'(x)=0$?

• First Derivative Test for inflection points

Answer 2

You are correct. $$f^{\prime}(x) = 2 + \ln(2) - 2x$$ Differentiating again we obtain $$f^{\prime \prime} = \frac{2}{x} - 2$$ Setting this equal to zero and solving we have $x = 1$. To test this, you can choose test points on the interval, say $x = \frac{1}{2}$ and $x = 2$. Notice that at $x = \frac{1}{2}$ the sign is positive and at $2$ the sign is negative. Notice that the domain of the function is $(0,1) \cup (1, \infty)$, hence why we chose those test points, as they are close to $1$ and in the domain. Since $\frac{1}{2}$ is positive we have that the function is concave up on the interval $(0,1)$ and concave down on on the interval $(1, \infty)$. Hence, the concavity of $f(x)$ changes from up to down. Hence, the only inflection point occurs at $x = 1$.

Answer 3

A point of inflection is one where $f$ changes from being convex to concave (or vice versa). If $f$ is differentiable, then $f$ is convex if and only if $f'(x)$ is increasing (and concave if and only if $f'(x)$ is decreasing). Thus at a point of inflection, $x_0$ say, $f'(x)$ switches from being increasing to decreasing or vice versa. The value $f'(x_0)$ at $x=x_0$ is thus unconstrained by the fact that $x_0$ is a point of inflection.

If $f$ is twice differentiable, then $f$ is convex if $f''(x)\geq 0$ (and concave if $f''(x)\leq 0$). It follows that at a point of inflection, if the second derivative is continuous, then $f''(x_0)=0$, and regardless, as long as $f''(x)$ is defined, $x_0$ is a point of inflection if and only if $f''(x)$ changes sign at $x=x_0$.

In the example you give of $f(x) = 2x\ln(x)-x^2$, $f'(x) = 2(1+\ln(x)-x)$ and $f''(x)= 1/x -1$, which is clearly positive for $x\in (0,1)$ and negative for $x \in (1,\infty)$ and hence $x=1$ is a point of inflection, but the sign of $f'(x)$ on either side of $x=1$ is not relevant.

Now consider a critical point $x_0$ of $f$, so that $f'(x_0)=0$. (What follows just expands on what is in user's answer to Do inflection points of $f(x)$ give $f'(x)=0$?). We will assume that $f$ is infinitely differentiable and that at our critical point $x_0$, we have $\{f^{(n)}(x_0): n\in \mathbb Z_{\geq 0}\}\neq \{0\}$, that is, the Taylor series of $f$ at $x_0$ is not identically zero. Let $k\geq 2$ be the smallest integer such that $f^{(k)}(x_0)\neq 0$. Then $f(x)$ is locally approximated near $x_0$ to order $k$ by $f(x_0)+(f^{(k)}/k!)(x-x_0)^k$.

Thus if $k$ is even then $x_0$ is a local maximum or minimum (and which of these holds is determined by the sign of $f^{(k)}(x_0)$). If $k$ is odd, then $x_0$ is not a local extremum, and $f''(x_0)=0$ (since $k\geq 3$). But then since, near $x_0$, $f''(x)$ is approximated to order $k-2\geq 1$ by $(f^{(k)}(x_0)/(k-2)!)(x-x_0)^{k-2}$ we see that $f''(x)$ changes sign at $x=x_0$ and hence $x_0$ is a point of inflection. Thus at least for infinitely differentiable functions whose Taylor series at the critical point $x_0$ do not vanish, the point $x_0$ is either a local extremum or a point of inflection.