Check for an onto function

Question

Why is $y= x^{2006} + x^{-2006} +5$ not an onto function if $f(x):\mathbb{R} \rightarrow \mathbb{R}$

Please provide an explanation.

Answer 1

if $\alpha>0$ then $\alpha+\frac{1}{\alpha}\ge\,2\,\,$. We know $x^{2006}\ge 0$ then $x^{2006}+\frac{1}{x^{2006}}\ge\,2\,\,$ as aresult $$y=x^{2006}+\frac{1}{x^{2006}}+5\ge\,2\,\,+5\ge7$$ i.e $R_f=[7,\infty)$

Answer 2

Hint: $x^{2006}$ is always non-negative. (Why?)

Furthermore, $\frac{1}{x^{2006}}$ is always positive when $x\neq 0$

(Technically, your proposed "function" is not a function at all since it is undefined for $x=0$. Regardless, looking at the domain instead as $\Bbb R\setminus\{0\}$ it will still have problems with being onto)

can you find a value of $x$ such that $f(x)=4$?

Answer 3

Onto means that $y = x^{2006} + x^{-2006}$ must hit every value in the range, i.e. $\mathbb{R}$.

But $x^{2006} > 0, x^{-2006} > 0,$ so clearly $y > 5$, and the range is $\mathbb{R}_{>5}$.