If a continuous function is strictly decreasing before a point and strictly increasing afterwards, is the point a global minimum? I'm in the middle of a proof that a point on a function is a global minimum. Usually I'd just solve an inequality to prove by contradiction that there are no points less than the minimum. But I can't in this case, since it's a transcendental equation (can't rearrange to make $x$ subject in terms of the elementary functions.) So I'm back to the drawing board. My question is:

If a continuous everywhere function is strictly decreasing before a point and strictly increasing afterwards, is the point a global minimum? How can this be proved?
Let us asume for contradiction that the given point is $x_0$, and that it is not a global minimum. Hence there is another point $x \neq x_0$ so that $f(x)&lt;f(x_0)$. But if $x \neq x_0$ then either $x&lt; x_0$ or $x&gt;x_0$, but since $f$ is strictly decreasing before $x_0$ then in the first case $f(x) &gt; f(x_0)$, in the second case similarly $f(x)&gt;f(x_0)$. We get a contradiction, and hence such $x$ does not exist. The continuity of $f$ is not necessary.
If you have been going downhill and at one stage things improved and never went down afterwards, then we say we turned around and that turning point was the worst. This is well-known to non-mathematicians too, does not calculus  techniques to be able to see this.