I answered your later question first, but this one follows the same kind of reasoning.
for a max slope we need the derivative of dy/dx = -4sin(x/2) which is
2cos(x/2)
setting that equal to zero ...
2cos(x/2) = 0
cos(x/2) = 0
x/2 = π/2, 3π/2 , 5π/2, and 7π/2
x = π, 3π, 5π, 7π etc
if x = 0, the slope is
-4sin(x/2)
= -4sin(π/2_) = -4
if x = 3π
slope = -4sin(3π/2) = +4
it will alternate thusly, so the max slope is +4
y=8cos((1/2)x)+10
dy/dx = -4sin((1/2)x)
What is the maximum slope from 0 - 4(pi)?
I tried equating the endpoints, but it does not work, can someone teach me how to find maximums and minimums?
1 answer