fastest increase rate is where the slope is steepest. That is, you want a maximum for f'(x). That happens where f"(x) = 0
f = 2cos(2x)
f" = -8cos(2x)
f"=0 when x=π/4, 3π/4 and so on.
The slope is negative at π/4, so you want x = 3π/4
A look at the graph confirms that the slope is greatest there.
http://www.wolframalpha.com/input/?i=2cos(2x)
How do you fund the fastest increase rate on a function like f(x) = 2cos(2x) ?
How about a rational function?
Thanks!
1 answer