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?
Best Answer
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