$f(x)=x(x^{2}-4)^{n}(x^{2}-x+1),n\in N$ assumes a local minima at $x=2$, find $n$

Question

$f(x)=x(x^{2}-4)^{n}(x^{2}-x+1),n\in N$ assumes a local minima at $x=2$, the $n$ has to be :

a) Even number

b) Odd number

c) Prime number

d) a Multipe of 4

Please help with this

I found $$f'(x)=(3x^{2}-2x+1)(x^{2}-4)^{n}+n(x^{2}-4)^{n-1}(x^{3}-x^{2}+x)=0$$

Any help on how to proceed will be highly appreciated.

Answer 1

$f(x)=x(x^2 - 4)^n (x^2 - x + 1)\quad (n\in N)$

$\text{$n$ even} \implies \begin{array}{|c|c|c|c|} \hline \text{function} & x=2-\epsilon & x=2 & x=2+\epsilon \\ \hline x & + & 2 & + \\ \hline (x^2-4)^n & + & 0 & + \\ \hline x^2-x+1 & + & 3 & + \\ \hline f(x) & + & 0 & + \\ \hline \end{array}$

$\text{$n$ odd} \implies \begin{array}{|c|c|c|c|} \hline \text{function} & x=2-\epsilon & x=2 & x=2+\epsilon \\ \hline x & + & 2 & + \\ \hline (x^2-4)^n & - & 0 & + \\ \hline x^2-x+1 & + & 3 & + \\ \hline f(x) & - & 0 & + \\ \hline \end{array}$

So $n$ needs to be even.

Answer 2

Notice that at $x = 2$, it is always the case that $f(2) = 0$. So it is sufficient to check the values of $f$ on the left and right of $x = 2$.

First, the term $x^2 - x + 1$ is always positive no matter the value of $x$ (you can check this by completing the square). So now suppose we only consider $g(x) = x(x^2 - 4)^n$.

One trick I remember is to draw a real line and mark the roots of the polynomial. Then you start filling alternatingly plus and minus signs (such as below, for example).

Since $x = 2$ is a local minimum, on the left of the first $2$ and on the right of the last $2$ should be the $+$ sign. You can see that in order to be so, $n$ has to be even.

Answer 3

We know that for positive integers $n$, $f(2) = 0$ trivially. Now, let's look at what happens in a sufficiently small neighborhood of $x = 2$, for the various factors of $f(x)$.

The factor $x$ behaves linearly, and is increasing in the neighborhood of $x = 2$.

The factor $x^2 - x + 1$ is quadratic but again, in a sufficiently small neighborhood of $x = 2$, it behaves like $y = (2(2)-1)(x-2) + (2^2 - 2 + 1) = 3(x-2) + 3 = 3x - 3$, so it too is increasing.

Where did this come from? What I did was compute the tangent line to $g(x) = x^2 - x + 1$ at $x = 2$; i.e., $$y = g'(2) (x - 2) + g(2).$$

So, all that remains is the factor $(x^2 - 4)^n$. When $x < 2$, we have $x^2 < 4$, hence the inside term is negative; when $x > 2$, we have $x^2 > 4$ and the inside term is positive. So if $n$ is even, the result is always nonnegative, and this factor exhibits symmetry about $x = 2$. But if $n$ is odd, the result is negative when $x < 2$ and positive when $x > 2$; i.e., it is antisymmetric. Since the other factors of $f$ are multiplicative and are locally linear around $x = 2$, these should not affect our conclusion. We should deduce that the correct answer is (a).

Answer 4

Hint: $\;f(x)=x(x-2)^{n}(x+2)^{n}(x^{2}-x+1)$ has a critical point at $x=2$ iff $n \ge 2$. The other factors $x\,$, $(x+2)^n\,$, $x^2-x+1$ are all positive and increasing at $x=2$ so the nature of the critical point is dictated by $(x-2)^{n}$ alone, which has a minimum at $x=2$ iff $n$ is even.

Answer 5

And one more:

$f(x) = (x^2 - 4)^n(x^3 - x^2 + x)$ ;

For $x = 2 $ second factor is $6$, and since continuos, remains positive in a small neighbourhood $ x$.

Behaviour of function in a small neighbourhood of $x = 2$:

$f(x) = A(x) (x^2 - 4)^n$, where $A(x)$ is the positive, bounded, second factor.

Let $n$ be odd:

$(x^2 -4)^n$ is negative for $x < 2$, implies $f(x)$ is negative.

$(x^2 - 4)^n$ is positive for $x > 2$, implies $f(x) $ is positive.

$f$ does not have a minimum at $x = 2$.

Let $n$ be even :

$(x^2 - 4)^n$ is positive for $x >2$ , or $x < 2$,

implies that $f$ is positive in a neighbourhood of $x =2$.

Hence a minimum.