I'm trying to prove the image of $f:\mathbb{R}\rightarrow\mathbb{R}\text{ by } f(x) = x^4+x^2$ is $\text{Im }f = \{x\in\mathbb{R}\ |x\geq0\}$ by cases considering $x>0,\;x<0,\;x=0$. And when considering the case $x>0$ is it okay to assume that $x^4+x^2>0$ since it is always positive?
If $x>0$ then $x^4+x^2>0$
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functions
elementary-set-theory
proof-verification
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0Yes, because the sum and the product of positive numbers are always positive. – 2017-02-15
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1It's unclear to me in what way you would "assume" that $x^4+x^2>0.$ I think rather for a proof at this level you would actually _show_ (in just a very few steps) that $x^4+x^2$ is always positive when $x>0.$ (And as ajotaxe points out, you can simplify this even further.) – 2017-02-15
1 Answers
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In fact, you needn't prove it by cases. $x^4$ and $x^2$ can't be negative, regardless of the sign of $x$, so their sum can't either.
Nevertheless, this only proves that $\text{Im }f \subseteq \{x\in\mathbb{R}\ |x\geq0\}$. You need also to prove that $f$ takes every positive value. Using continuity and intermediate value theorem should work.