Question

how do i find the absolute extremum of t sqr(4-t^2) [-1,2]

Answer 1

Hi there!
\(\Large \color{royalblue}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

\[t \sqrt{4-t^2} \left[ -1,2 \right]\]

Answer 3

Hi

Answer 4

I guess to start, we would want to make sure that our function is `continuous` over the given interval.

So in this case, our function has a domain of:\[\Large t\in [-2,2]\]So we don't have any problems there, good good.

Answer 5

dont i have to sqr both sides and then take the derivative if each side

Answer 6

\[\Large f(t)\quad=\quad t\sqrt{4-t^2}\]To find extrema, we'll take the derivative and set it equal to zero.\[\Large f'(t)\quad=\quad ?\]

Mmm no let's not square each side! That will cause a big mess for us.
Looks like we'll want to apply the product rule for this one.

Answer 7

\[\Large f'(t)\quad=\quad \color{royalblue}{(t)'}\sqrt{4-t^2}+t\color{royalblue}{\left(\sqrt{4-t^2}\right)'}\]There is our setup for product rule.
We need to take the derivative of the blue pieces.
Make sense so far? :o

Answer 8

right
so it would be \[1\sqrt{4-t^2}+t(2t)\]

Answer 9

is that right

Answer 10

Hmm 2t?\[\Large \frac{d}{dx}\sqrt{4-t^2}\quad=\quad \frac{d}{dx}(4-t^2)^{1/2}\]Applying the power rule:\[\Large \frac{1}{2}(4-t^2)^{-1/2}\color{orangered}{\frac{d}{dx}(4-t^2)}\]We have to apply the chain rule, that's where the orange part is coming from:\[\Large =\frac{1}{2}(4-t^2)^{-1/2}\color{orangered}{(-2t)}\]If we simplify it down a bit, and rewrite our 1/2 power as a root,\[\Large =\frac{-t}{\sqrt{4-t^2}}\]

Answer 11

ok so now all i do is substitute -1 and 2 in for t right

Answer 12

So we found our derivative:
\[\Large f'(t)\quad=\quad \sqrt{4-t^2}+\frac{-t^2}{\sqrt{4-t^2}}\]Now we want to set the derivative equal to zero and solve for t to find critical points.\[\Large 0\quad=\quad \sqrt{4-t^2}+\frac{-t^2}{\sqrt{4-t^2}}\]

Answer 13

Start by getting a common denominator:\[\Large 0\quad=\quad \frac{(4-t^2)-t^2}{\sqrt{4-t^2}}\]

Answer 14

how did you get t^2 on top

Answer 15

i know that the common denominator is 4-t^2

Answer 16

\[\Large f'(t)\quad=\quad \color{royalblue}{(t)'}\sqrt{4-t^2}+t\color{royalblue}{\left(\sqrt{4-t^2}\right)'}\]So our derivative gave us this:\[\Large f'(t)\quad=\quad \color{orangered}{(1)}\sqrt{4-t^2}+t\color{orangered}{\left(\frac{-t}{\sqrt{4-t^2}}\right)'}\]In the second term there, I brought the black t into the numerator and multiplied it by the -t.

Answer 17

ok i see now .... i missed that "t"

Answer 18

Woops I forgot to take the prime off of the orange block :) it shouldn't be there anymore.

Answer 19

right i figured that...

Answer 20

Normally, to solve for t, we would set the numerator equal to zero.
To find critical points we will also have to consider places where the derivative might be undefined.

For our problem that means we need to also set the denominator equal to zero.\[\Large 0\quad=\quad \sqrt{4-t^2}, \qquad\qquad\qquad 0\quad=\quad (4-t^2)-t^2\]So we need to solve for t each in each of these cases.

Answer 21

What values do you get for t? :)

Answer 22

-1,2

Answer 23

Hmmm...

Answer 24

For this one:\[\Large 0\quad=\quad \sqrt{4-t^2}\]Squaring each side:\[\Large 0\quad=\quad 4-t^2\]Solving for t gives us:\[\Large t\quad=\quad \pm \sqrt4\qquad\to\qquad t\quad=\quad\pm2\]

Answer 25

right that is what i got for that one

Answer 26

For the numerator we should get something similar: 2 answers due to the square.

Answer 27

\[\Large 0\quad=\quad(4-t^2)-t^2\]\[\Large 0\quad=\quad 4-2t^2\]

Answer 28

Hmm I don't think it'll give us -1 :O try again maybe? :D

Answer 29

ok \[4-2t^2= 0 \rightarrow 2t^2=-4 \rightarrow 2t^2\div2 =4\div2\]

Answer 30

that is how i am trying to solve this...am i doing it wrong?

Answer 31

\[\Large t^2\quad=\quad 2\]Nope, looks good so far!

Answer 32

ok so then t^2=2 would be 1 right

Answer 33

if you squared it

Answer 34

taking the square root of each side gives us:\[\Large t\quad=\quad \pm\sqrt2\]

Answer 35

ok so our critical numbers are \[\pm 2,\pm \sqrt{2}\]

Answer 36

Yes,\[\Large t=-2,\;-\sqrt2,\;\sqrt2,\;2\]Recall that our interval is \(\Large t\in[-1,2]\)
Do all 4 of our critical points fall inside of this interval?

Answer 37

no

Answer 38

Hmm so we'll have to exclude any values that don't fall in our interval.
Which I guess ends up leaving us with:\[\Large t=\sqrt2,\;2\]

Answer 39

right

Answer 40

so now we take our critical numbers and plug in

Answer 41

So now comes the easier part.
We have our `critical points` and our `end points`.

We plug all of those values into the function f(t) and compare.
The largest output is our max, the smallest our min.

Answer 42

\[\Large f(-1)\quad=\quad ?\]\[\Large f(\sqrt2)\quad=\quad?\]\[\Large f(2)\quad=\quad ?\]

Answer 43

so we plug into the original problem not the derivative...right

Answer 44

ok so i got f(-1)=-2.24 f(\[\sqrt{2} =2\] f(2) = 0

Answer 45

Mmm ok good!
So x=-1 gave us the smallest value
and x=sqrt2 gave us the largest value.

So we've found our max/min points within the given interval!

Answer 46

Write max/min as ordered pairs.

Answer 47

max (\[\sqrt{2}\],2)

Answer 48

min (-1,-2.24)

Answer 49

yay good job \c:/