Calculus questions...PLEASE READ...?

Question

Okay, we are going over application of derivatives in class right now...I missed the lecture and am trying to understand this stuff the best I can off some notes that I borrowed...It's all about extrema...

1. The veolocity of a particle is (ft/s) is given by v=t^2 -7t +9, where t is time for which it has traveled.Find the time at which the velocity is at a minimum...
(I got the answer 3.5 s. Please correct if wrong...)

2. Find the absolute extreme values of each f(x).
f(x)= cscx [-pi/2 , 3pi/2]
(Would a max. and min. even exist?...This one really confuses me...Please help if you can...)

And lastly
3. Find the derivative at each critical point and determine the local extreme values...
(I don't really understand what this one is asking...)

PLEASE HELP IF YOU CAN...

Answer 1

1. find the derivative of dv/dt. Find t when dv/dt = 0

2. find df/dx. Find the values of x that make df/dx = 0.

3. What is the function? You probably need to solve y' = 0 (and pehaps y"=0 depending on the function).

Answer 2

1. The veolocity of a particle is (ft/s) is given by v=t^2 -7t +9, where t is time for which it has traveled.Find the time at which the velocity is at a minimum...

I got the answer 3.5 s. Please correct if wrong... <---- your answer is correct

2. Find the absolute extreme values of each f(x).

f(x)= cscx [-π/2 , 3π/2]
= 1/sinx [-π/2 , 3π/2]

Now sin x = 0 when x = 0, π
Thus csc x has discontinuities at x = 0, π

So lim x→0- = -∞ and lim x →0+ = +∞
and lim x→π- = +∞ and lim x →π+ = -∞

Other critical points
Min sin x = -1 at -π/2, 3π/2 and max sin x = +1 at x = π/2
So cscx = -1 at -π/2, 3π/2 and csc x = +1 at x = π/2

And lastly
3. Find the derivative at each critical point and determine the local extreme values...

f'(x) = - cscx cotx
= 0 for stationary points
ie 1/sinx * cosx/sinx = 0
so cos x = 0 ie x = -π/2, π/2, 3π/2

f''(x) = cscxcot²x + csc³x
When x = -π/2, 3π/2, f''(x) = -1 < 0 so curve concave down. So local maximum at (-π/2, -1) and (3π/2, -1)
When x = π/2, f''(x) = 1 > 0 so curve concave up. So local minimum at (π/2, 1)

Answer 3

well. sorry. i can only help u up to the 1st qn.
u r correct. for velocity at minimum, derivative = 0.
so 2t - 7 = 0

(for further understanding)
as u noe, the derivative of a function is the gradient function, so if a graph or v/ ft/s against t/s is drawn, the pt where it is a minimum will be a turning point, and at that point the gradient is zero since the tangent is a horizontal line.
so the derivative = 0.