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.

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.
