find the absolute max and abs min?

Question

find the abs max and abs min values of f on the given interval..

f(t)= cuberoot(t) (8-t), [0,8]

please explain what you did step by step.. in like an easy to understand way.. (sorry.. im really tired and stupid right now)

Answer 1

min of f(t) = 0 when t = 0 and t = 8
max occurs at t=2

I assume you mean the maximum and minimum of the absolute value of f(t). If that's the case then the minimum of f(t) occurs when t= 0 or 8 where f(t) =0

To determine the max we find when the derivative f ' (t) = 0

let g(t) = cuberoot(t) = t^(1/3) and h(t) = (8-t)

The derivative of f(t) (i.e. f ' (t)) is given by

f ' (t) = g'h+h'g (the product rule)

f ' (t) = (8-t)(1/3)t^(-2/3) - t^(1/3)

The max, min and inflexion points occur when f ' (t) =0

t^(1/3) = (8/3)t^(-2/3) - (1/3)t^(1/3)

(4/3)t^(1/3) = (8/3) t^(-2/3)

t =2

f(2) = 7.56 apprx is the max

Answer 2

easy way is to use a calculator of course...most graphing calculators have a function that finds the max and min for you..but if you must show your work then here is how it is done====

1. Find the first derivative of f(t) using product rule==
f'(t)=(8-4t)/(3t^(2/3))
2. Where f'(t)=0 is a horizontal tangent line===
(8-4t)/(3t^(2/3))=0
note* this is equal to zero when the top is equal to zero
thus==== 8-4t=0 t=2

3. Now with this info all you need to do is plug this value and the ends of the interval into the preliminary equation==
t=0 f(t)=0
t=2 f(t)=6(2^(1/3))
t=8 f(t)=0

4. therefore the abs mins are (0,0) and (8,0)
abs max is (2, 6(2^(1/3))

Answer 3

Did you mean t^(1/3) * (8-t)?

If so, take the derivative and set it equal to zero. You test the t values where the derivative is equal to zero and you test the endpoints. The t value where the function is largest is the absolute maximum and the t value where the function is smallest is the absolute minimum.

I see you wanted step by step:

Distribute the t^(1/3) over (8-t) and get

8 t^(1/3) - t^(4/3)

Take the derivative:

(8/3) t^(-2/3) - (4/3) t^(1/3)

Set it equal to zero

(8/3) t^(-2/3) - (4/3)t^(1/3) = 0

(8/3)t^(-2/3) = (4/3) t^(1/3)

t = 2

f(0) = 0
f(2) = 2^(1/3) * 6
f(8) = 0

Answer 4

f(t) = cuberoot(t)(8 - t)

Easiest way to do this: graph this on a calculator and see where the function is the highest between t=0 and t=8. Or, look online for a free graphing program. When I graphed it, f(t) is lowest at t=0 and 8 (y coord. is 0) and highest at about t=1.8 (the y coord. is about 7.5)

Answer 5

once you plug in -one million and then 3 into the equation 5x-one million it provides 14 and -6 with the aid of fact the max and min. often you will ought to plug those into the 1st spinoff yet with the aid of fact the 1st spinoff is 5 and not an equation right here you do no longer ought to try this.

Answer 6

cube root of f(t) can be rewritten as (8-t)^(1/3) = f(t)

Take the derivative dy/dt = -1/3 (8-t)^(-2/3)

We see that in [0,8] this derivative is always negative (8-t) is always positive on that interval

for t=0 f(t) = cube root of 8 which is 2

fot t=8 f(t) = cube root of 0 which is zero

Answer 7

use a calculator! if not then try to plug in the values, and see where it is the highest or lowest

Answer 8

Look at the left boundary and the right one !!

Th