Find the minimum value of $t^3+t^2-2t-2$ given that $t$ is greater than or equal to 2 The original question was to find the range of the function f defined by:
$$f(x)=\frac {(1+x+x^2)(1+x^4)}{x^3}$$ for $x>0$
Evidently, differentiating is not very helpful. So I wrote $f(x)$ as:
$$f(x)=t^3+t^2-2t-2=(t^2-2)(t+1)$$
Where $t=x+\frac 1x$
The maximum value clearly approaches $+\infty$.
How do I find the minimum value, where $t >=2$? Here, too differentiating would not help, unless I want to plot a graph, which would again be somewhat cumbersome.
 A: The following is your solution, and is very nice.  Note that our function is equal to 
$$\frac{1+x+x^2}{x}\cdot \frac{1+x^4}{x^2},$$
which is 
$$\left(x+1+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right).$$
Make the substitution $x+\frac{1}{x}=t$. Our function is equal to 
$$(t+1)(t^2-2).\tag{1}$$
Since $t\ge 2$, the function (1) is increasing, so has a minimum value of $6$ at $x=1$.  
A: I woud say that
$\left(\frac{(1+x+x^2)(1+x^4)}{x^3}\right)' = \frac{3 x^6+2 x^5+x^4-x^2-2 x-3}{x^4}$
The equation $3 x^6+2 x^5+x^4-x^2-2 x-3=0$ is a negatively reciprocal equation of odd degree, obviously has roots
$x=1, x=-1 \Rightarrow 3 x^6+2 x^5+x^4-x^2-2 x-3 = (x-1)(x+1)(3 x^4+2 x^3+4 x^2+2 x+3) = 0$
The equation $3 x^4+2 x^3+4 x^2+2 x+3=0$ is a positively reciprocal equation, solution
after adjusting the shape dividing $x^2$, substitution $t = x + \frac{1}{x} \Rightarrow$ quadratic equation, this has no real roots. 
So may be extreme at points x = -1 or x = 1.
I am sure that you can take from here.
A: Expanding $f(x)$ and breaking the fraction,
$$f(x)=\frac{(1+x+x^2)(1+x^4)}{x^3}=\frac{1+x+x^2+x^4+x^5+x^6}{x^3}=\frac 1{x^3}+\frac 1{x^2}+\frac 1x+x+x^2+x^3.$$
Can you differentiate $f(x)$ now? Only the power rule is required for the last expression.
