you have found the variables which will yield the maximum and minimum , but you have not found the actual max or min
When t = 0 (you had x = 0)
h(0) = 0-0 = 0
when t = 2, h(2) = 8 - 12 = -4
also h(-1) = -1 - 3 = -4
so you have a local max at (0,0) and a local min at (2,-4)
for the given domain, there is a relative min at (-1,-4) as well
(I do not know how your text or your teacher defines "relative and absolute extrema"
for g(t) = 3t^4 - 16t^3 + 24t^2 + 1
g ' (t) = 12t^3 - 48t^2 + 48t
= 0 for max/mins
divide by 12
t^3 - 4t + 4 = 0
nasty equation to solve, I used Wolfram to get
t = appr -2.383
http://www.wolframalpha.com/input/?i=solve++x%5E3+-+4x+%2B+4+%3D0
plug in that value to find the g(t) value.
looking at the graph of g(t) shows me it is a minimum.
f(x) = 3x^4 - 2x^3
f ' (x) = 12x^3 - 6x^2 = 0 for max/min
divide by 6
2x^3 - x^2 = 0
x^2(2x - 1) = 0
x = 0 or x = 1/2
hope that helps
Please tell me if this are right. Thank you!
Find the exact location of all the relative and absolute extrema?
h(t) = t^3 - 3t^2 domain: [-1, + infinity)
For this one I got x=2 as the minimum, and x=0 as the maximum.
g(t) = 3t^4 - 16t^3 + 24t^2 +1 domain: (-infinity, +infinity)
For this one I got x=0
f(x) = 3x^4 -2x^3 domain: [-1, + infinity)
I'm not sure about this one I think it is x=0
1 answer