Find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results.

h(t)=(t-1)^2/3
[-7,2]

Question

Find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results.

h(t)=(t-1)^2/3 
[-7,2] <--interval

anonymous · Answer

For the absolute minimum I got (-7,-4).
For the absolute maximum I got (2,1).

The absolute maximum shown in the book is (-7,4)..did i do something wrong?

rogue · Answer

Well, your absolute min is wrong as well ;)

You can figure out the min and max from thinking about how the graph looks, but since this is calculus, I guess you have to find them via differentiation :P

What did you get for the first derivative of h(t)?

anonymous · Answer

see i don't know why my teacher does the derivative when i can find the Critical numbers simply through the original function. 

But it's 2/3(t-1)^-2/3

rogue · Answer

Yeah, that's correct. Now the critical points of the function are where the derivative equals zero or doesn't exist. Where does that occur?

rogue · Answer

the power of (t-1) in the derivative should actually be -1/3

anonymous · Answer

no it's -2/3 b/c it's minus 1 when you do product rule...and it's 1.

rogue · Answer

$$\frac {2}{3} - \frac {3}{3} = \frac {-1}{3}$$ hehe ;D

anonymous · Answer

oh right -_-

rogue · Answer

Anyways, the derivative is

$$h(t)= \frac {2}{3}(t−1)^{−1/3}=\frac {2}{3(t−1)^{1/3}}$$

There are no instances where the derivative equals 0, but at t = 1, the derivative is undefined, so that is a critical point. Can you go on from there?

anonymous · Answer

for the most part yes, but the interval is [-7,2], what do you get when you substitute -7 in the original equation? What do you get for h(t)?

rogue · Answer

h(-7) = (1 - -7)^(2/3) = 8 ^ (2/3) = cube root of 8^2 = 4

anonymous · Answer

but it's t-1, not 1-t...so (-7-1)^(2/3)=-4 correct?

rogue · Answer

oh, yes, but either way, it's positive 4.

cube root of (-8)^2 = cube root of 64 = 4
(cube root of -8)^2 = (-2)^2 = 4

anonymous · Answer

ah ok. thanks! my calculator was giving me -4 for some reason...

anonymous · Answer

and it still is...

anonymous · Answer

nvm, parenthesis issue -_-

rogue · Answer

If you plug in the critical value of t = 1, you'll get the absolute min of 0 btw.