# Calculus- Find the maximum height of a function

A ball is thrown into the air with an initial velocity of $16 ft/s$. its height after $t$ seconds is given by $f(x) = 16t-4t^2$ . After how many seconds does the ball reach its maximum height?

I can't remember back whether this fell under the derivative section or the Limit section, therefore i am stuck

• Are you sure the equation is right? From what I see its max will be at t=0 – Matt Dec 2 '15 at 2:07
• Sorry, I had a - infront of the 16. my bad. – Jon Roy Dec 2 '15 at 2:09
• @Matt - You may be right, however i got two. – Jon Roy Dec 2 '15 at 2:16
• Yeah, the formula changed, in the first version it was just 16, not 16t - that or I missaw. 2 is right, get it by setting the derivative to 0 – Matt Dec 2 '15 at 2:17

## 1 Answer

My advice is not to think of problems as being in "sections", but instead to think about what the problem is asking for.

That function is fairly well-behaved - no weird corners or jumps or holes.

So, where it reaches its maximum, it's going to be flat before falling back down, right?

HINT: If it's flat at its maximum what does that tell you about it?

• I calculated the maximum at x = 2, using the increasing/decreasing interval method, seem correct? – Jon Roy Dec 2 '15 at 2:12
• @Jon: You got it! – Deusovi Dec 2 '15 at 2:20