Question

Find the minimum of f(x)=xsqrt(x+1) on the interval (-1,1)

Answer 1

I got (uv)'=u'v+uv'
u=x
u'=1
v=sqrt(x+1)
v'=1/2(x+1)^(-1/2)

Answer 2

with that i got
x(1/2)(x+1)^(-1/2)+sqrt(x+1)
now idk what to do??

Answer 3

you got the v' and u' right.

Answer 4

yes, u'=1 and v'=(1/2)(x+1)^(-1/2)

Answer 5

and the (uv)' together is correct also.

Answer 6

\[v'=\frac{ 1 }{ 2 }(x+1)^{\frac{ -1 }{ 2 }}\]

Answer 7

so i set it to 0?

Answer 8

yes, you set (uv)' equal to zero.

Answer 9

\[x \frac{ 1 }{ 2 }(x+1)^{\frac{ -1 }{ 2 }}+\sqrt{x+1}=0\]

Answer 10

\(\large\color{black}{ 0=\frac{x}{2\sqrt{x+1}} +\sqrt{x+1} }\)

Answer 11

I would play with the (uv)' for a bit though.

Answer 12

should i devide both sides by \[\sqrt{x+1}\]?

Answer 13

okay

Answer 14

\(\Large\color{black}{ 0=\frac{x}{2\sqrt{x+1}} +\frac{2(x+1)}{2\sqrt{x+1}} }\)

\(\Large\color{black}{ 0=\frac{x}{2\sqrt{x+1}} +\frac{2x+1}{2\sqrt{x+1}} }\)

\(\Large\color{black}{ 0=\frac{3x+1}{2\sqrt{x+1}} }\)

Answer 15

See what I did?

Answer 16

now, multiply both sides times \(\Large\color{black}{ 2\sqrt{x+1} }\) .

Answer 17

and, ... Are you sure that the interval is (-1,1) and NOT [-1,1] ?

Answer 18

on the hw sheet it has -11,1 in () as (-1,1)

Answer 19

but I am following what you have done so far

Answer 20

So your interval is \(\Large\color{black}{ (-11,1) }\) ?

Answer 21

and im left with 3x+1)=0

Answer 22

yes, with 3x+1=0, so x equals?

Answer 23

u there ?

Answer 24

1/3=x

Answer 25

-1/3, no?

Answer 26

yes sorry, scrolled up for a second
oh yes -1/3

Answer 27

and again, your interval is (-11,1) correct?

Answer 28

(-1,1) not -11

Answer 29

Ohh. if it is \(\Large\color{black}{ (\color{red}{-1},1) }\), then -1, is not a critical number, even though it is in the domian of f(x), and f'(x) is undefined at it.

Answer 30

Then, -1/3 would be your only critical number.

Answer 31

would finding f'' get me the minimum then?

Answer 32

No, you needed the critical numbers. And the smallest f ' (critical number) you get, that is the the absolute minimum.

Answer 33

f '' (x) has to do with testing concavity, and it is different.

Answer 34

okay! So than -1/3 is the only answer I need than. You are awesome thank you soo much!!!!

Answer 35

You are not done.

Answer 36

-1/3 is the critical number with which you would get an absolute minimum of the function,
but the absolute minimum itself, is f(-1/3).

Answer 37

Calculate f(-1/3)

Answer 38

okay!!!
f(-1/3)=1/3sqrt(4/3)

Answer 39

\[f(\frac{ -1 }{ 3 })=(\frac{ 1 }{ 3 })\sqrt{\frac{4 }{ 3 }}\]

Answer 40

can that be simplified?

Answer 41

it is negative 1/3, not positive 1/3.

Answer 42

oh
yes

Answer 43

it can
\[f(\frac{ -1 }{ 3 })=(-\frac{ \sqrt{\frac{ 4 }{ 3 }} }{ 3 })\]

Answer 44

ohhh i see it is negative
than that means the 4/3 in sqrt is 2/3