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
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
To find the extrema of a function, you need to take the derivative and solve for where the derivative equals zero. The critical points obtained from this process can be potential extrema. In addition, you should also check the values of the function at the endpoints of the given domain.
Now let's apply this method to each given function to determine the exact locations of the extrema:
1. Function h(t) = t^3 - 3t^2, with domain [-1, +infinity)
To find the critical points, we first take the derivative of h(t): h'(t) = 3t^2 - 6t.
Setting h'(t) equal to zero and solving for t:
3t^2 - 6t = 0
t(3t - 6) = 0
From this equation, we find two potential critical points: t = 0 and t = 2.
Next, we check the values of the function at the endpoints of the domain: h(-1) and as t approaches infinity.
h(-1) = (-1)^3 - 3(-1)^2 = -1 + 3 = 2
As t approaches infinity, h(t) also approaches infinity.
Based on these calculations, we find that the function h(t) has two potential extrema:
- Minimum at t = 2
- Maximum at t = 0
2. Function g(t) = 3t^4 - 16t^3 + 24t^2 + 1, with domain (-infinity, +infinity)
First, we find the critical points by taking the derivative of g(t): g'(t) = 12t^3 - 48t^2 + 48t.
Setting g'(t) equal to zero and solving for t:
12t^3 - 48t^2 + 48t = 0
12t(t^2 - 4t + 4) = 0
12t(t - 2)^2 = 0
We find one critical point from this equation: t = 0.
Given the domain (-infinity, +infinity), we do not have any endpoints to consider.
Therefore, the function g(t) has one potential extremum at t = 0.
3. Function f(x) = 3x^4 - 2x^3, with domain [-1, +infinity)
To find the critical points, we take the derivative of f(x): f'(x) = 12x^3 - 6x^2.
Setting f'(x) equal to zero and solving for x:
12x^3 - 6x^2 = 0
6x^2(2x - 1) = 0
From this equation, we find two potential critical points: x = 0 and x = 1/2.
Next, we check the value of the function at the endpoint of the domain: f(-1).
f(-1) = 3(-1)^4 - 2(-1)^3 = 3 - 2 = 1
Based on these calculations, we find that the function f(x) has two potential extrema:
- Minimum at x = 0
- No maximum within the given domain
Please note that these results may be subject to verification and further analysis, but the described process should help you verify the locations of relative and absolute extrema for each function.