I'm in college calc class, so please keep explanations simple.
Let $g(t)=\sqrt{1-\frac{1}{t+1}}$.
I've established its domain to be $(-\infty,-1) \cup [0,\infty)$.
I'm not sure how to go about determining boundedness, please provide feedback on my method below. My understanding is that a function is unbounded if it can output arbitrarily large positive or negative values.
My strategy has been to look at what happens to the function when x is close to the edges of the domain. For example if $x=-1.01$ then $\frac{1}{t+1}$ becomes a "large negative value" -10 and thus the whole function becomes $\sqrt{1-{LargeNegativeNumber}}$ and thus on the whole we get $\sqrt{LargePositiveNumber}$ which is unbounded upwards (in the positive y direction).
If we look at another edge of the domain, $ -\infty $, we see that $\frac{1}{-\infty+1}$ becomes a small negative value. Thus the whole function becomes $\sqrt{1-{SmallNegativeNumber}}$ thus it approaches $1$ from above. (This dcheck doesn't alter our understanding of whether g is bounded or not, right?)
Ok let's go on to looking at values close to 0, say $x = 0.01$. In that case $\frac{1}{0.01+1}$ becomes a "large negative value" and the whole function becomes $\sqrt{1-{LargeNegativeNumber}}$ and thus $\sqrt{LargePositiveNumber}$. Ok this confirms our understanding that g is unbounded upwards in the positive y direction.
Ok then I look at what ahppens to the function when $x = \infty$. We see that $\frac{1}{\infty+1}$ becomes a "small positive number" and thus the whole function becomes $\sqrt{1-{SmallPositiveNumber}}$ and thus 1 approached from below.
My conclusion is that the function is unbounded upwards, but the answer says it's just "not bound" which I assume means unbounded upwards and downwards, but I've failed to show unboundedness downwards! What did I do erroneously?
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