Existence of extremal value

Question

Determine the minimum and maximum values of the function $\frac{x^4}{(x-1)(x-3)^3}$. By equating the first derivative to $0$,its fairly simple to deduce that the extreme points are $x=0,\frac{6}{5}$. However,to determine minimum or maxumum,we need to analyse the 2nd derivative as well which seems to be awfully long. Is there any way to do it simply?

Answer 1

Call your function $f$. Notice that the numerator is always non-negative. Now, in some neighbourhood of $0$ we have $x-1<0$ and $x-3<0$, so the denominator is always non-negative in that neighbourhood. As $f(0)=0$, it must be a minimum.

As the only roots of $f'$ are $0$ and $\dfrac{6}{5}$ and it is a continuous function, it must have a constant sign on $\left(1,\dfrac{6}{5}\right)$ and a constant sign on $\left(\dfrac{6}{5},3\right)$. Evaluating, one gets that $f'\left(\dfrac{11}{10}\right)>0$ and $f'(2)<0$ , so $f$ increases on $\left(1,\dfrac{6}{5}\right)$ and decreases on $\left(\dfrac{6}{5},3\right)$, from where $f$ attains a maximum at $\dfrac{6}{5}$.

Answer 2

The function tends to $1$ at $-\infty$ and to $\infty$ when $x\to 1^-.$ As $0$ is the only stationary point in $(-\infty,1),$ it must be a minimum. On the other hand $6/5$ belongs to the interval, where the function tends to $-\infty$ at $1^+$ and at $3^-.$ Therefore the function attains a maximum at $6/5 $