finding upper bounds I am currently working on analysis papers and i came across several questions involving supremum or infimum.
I know that the supremum is the least upper bound and the infimum is the highest lower bound.
However, I cannot even find a starting point to answer these questions involving bounds.
For example, 
Let a < b be real numbers.
Suppose that f:(a,b) -> R is differentiable.Let c be element of (a,b).
If f(x)=(x^4)-(x^2), 
prove that f is bounded below. Is f bounded above? justify
In a previous part of the question, I found the critical points of f.
Please help me as this is a very important notion and is involved in the Extreme value theorem.

Note that $f(x) = 0$ precisely at the points $x=-1, 0, 1$. Since $f(2) = f(-2) = 12$ and $f$ is continuous, we must have $f(x) > 0$ for all $x > 1$ and $x < -1$, so $f$ is bounded below by $0$ on these intervals.
Furthermore, the interval $[-1,1]$ is compact (closed and bounded), so again by continuity, $f$ is bounded above and below on this interval.
So we have established that $f$ is bounded below on all three intervals $(-\infty, -1)$, $[-1,1]$, and $(1, \infty)$. Therefore it is bounded below, by the minimum of the lower bounds on these three intervals.
On the other hand, $f$ is clearly not bounded above, because $\lim_{x \to\infty}f(x) = \infty$.
Note that we did not need differentiability in order to reach this conclusion. Also, although you mentioned infimum and supremum in your first two sentences, the question itself does not seem to involve them.
To compute the infimum and supremum, you can take advantage of the differentiability of $f$, which ensures that any local extrema must occur at critical points. The critical points are where $f'(x) = 0$. In this case, $f'(x) = 4x^3 - 2x$, which equals zero if and only if $x(4x^2 - 2) = 0$, if and only if $x = 0$ or $x = \pm 1/\sqrt{2}$. Evaluating $f(x)$ at each critical point, we obtain $f(0) = 0$ and $f(\pm 1/\sqrt{2}) = -1/4$. Since $f$ is bounded below, its infimum must be the smallest of these values, namely $-1/4$. As we noted earlier, $f$ is not bounded above, so its supremum is $+\infty$.
