This is the problem I have.
"Find all critical points in the following problem.
k(t)=5/square root[t-squared+1]"
2006-08-31 10:26:45 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
This is Calculus? I did this in my Advanced Algebra class... I'm confuzzled.
2006-08-31 13:45:01 · answer #1 · answered by jpklla 3 · 0⤊ 2⤋
A critical point is point of the function. The x value must be root or discontinuity of first derivative.
So:
1) find the derivative of the function
it is: y´ = -5t/sqrt((t^2+1)^3)
2) there are no discontinuity value... and there is only a root t=0
3) Then the critical point is (0,5)... it is the only probably point of a maximum or a minimum
2006-08-31 17:46:30 · answer #2 · answered by vahucel 6 · 0⤊ 0⤋
sqrt (t^2 + 1) will have a minimum value at t = 0
and max is infinity
as at all other values, t^2 is positive...
so, k(t) = 5/ sqrt (t^2 + 1)
has a max value of 5/ 1 = 5
and minimum value approaches 0... asymptodic curve
so, the curve will look like a normal distribution curve or a bell curve with max at t = 0 and value will be 5
the only critical point is t = 0
2006-08-31 17:34:57 · answer #3 · answered by DG 3 · 0⤊ 0⤋
since k'(t) exists for all reals, the only thing you have to do is solve k'(t)=0 and solve for t. you should get that the only critical point is t=0.
2006-08-31 17:37:38 · answer #4 · answered by a_liberal_economist 3 · 0⤊ 0⤋
Just checked in to see the answer. So many to choose from.
2006-08-31 18:23:21 · answer #5 · answered by NoPoaching 7 · 0⤊ 0⤋
dude, just use your calculator to find the max and mins
2006-08-31 17:30:35 · answer #6 · answered by Giovanni McAdoo 4 · 0⤊ 0⤋
find the derivative, set it to zero, and solve that equation.
2006-08-31 17:40:54 · answer #7 · answered by anonymous 3 · 0⤊ 0⤋