
Distance formula
Distance: $\displaystyle s=16t^2+800t$
find t, for s increasing, for s'=0, and for s decreasing
Am I figuring maximums and minimums here?
$\displaystyle y'=32t+800$ =$\displaystyle y'=t+25$
so I have a critical point at x=25
$\displaystyle y''=1$
So our singular critical point is a maximum. So t>25 we start to decrease?
I have a multiple choice solution set
a.) (0,25]
increasing
{0,50}
s=0
b.)[25,50]
decreasing
{0,50}
s=0
c.) {0,50}
s=0
d.) None of these
Keep in mind. He likes to trick us and I've noticed a lot of "none of these" resulting from things like positive and negative signs in the equation. Part of the thing confusing me is how he wrote up the solution set

As you probably know, it is a concave down parabola with vertex at
V(25,10000).
It follows that $\displaystyle f'(x) < 0 \;\ if \;\ 25 < x < {\infty}$
and $\displaystyle f'(x) > 0 \;\ if \;\ {\infty}<x<25$
Since f(x) is continuous at x=25:
Therefore it is decreasing over the interval $\displaystyle [25,\infty)$ and increasing over the interval $\displaystyle (\infty, 25]$
$\displaystyle f''(x) < 0>f''(x) = 32$, so it is concave down. We can tell that from the negative coefficient of x^2.