g(x) = e^f(x)
g'(x) = e^f(x) f'(x)
g'(1) = e^f(1) f'(1) = e^2 (-4) = -4e^2
g(1) = e^f(1) = e^2
so, you want the line through (1,e^2) with slope -4e^2:
y-e^2 = -4e^2 (x-1)
g has a max where g' = 0
since e^f > 0 for all x, g'=0 when f'=0. So, g has a max/min at x = -1 or x=3
Since f''<0 at x=-1, g is a max at x = -1.
e^f > 0
f'(-1) = 0
f''(-1) < 0
so, g''(-1) < 0
g'(3) = e^f(3) f'(3) = 0
g'(1) = -4e^2 as above
avg change is
(g'(3)-g'(1))/(3-1) = (0- -4e^2)/2 = 2e^2
Two particles move along the x -axis. For 0 is less than or equal to t is less than or equal to 6, the position of particle P at time t is given by p(t)=2cos((pi/4)t), while the position of particle R at time t is given by r(t)=t^3 -6t^2 +9t+3.
1. For 0 is less than or equal to t is less than or equal to 6, find all times t during which particle R is moving to the left. 2. for 0 is less than or equal to t is less than or equal to 6, find all times t during which the two particles travel in opposite directions. 3. Find the acceleration of particles P at time t=3. Is particle P speeding up, slowing down, or doing neither at time t=3? Explain your reasoning. 4. Show that during the interval (1,3), there must be at least one instant when the particle R must have a velocity of -2.
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