Question

Let f(x)= √(x-1). Find all the values of x in the interval (1,5) guaranteed by the mean value theorem

Answer 1

you need a couple numbers first

Answer 2

we need to compute
\[\frac{f(b)-f(a)}{b-a}\] for \(f(x)=\sqrt{x-1},a=1,b=5\)

Answer 3

i get
\[f(1)=\sqrt{1-1}=0,f(5)=\sqrt{5-1}=\sqrt{4}=2\] and so
\[\frac{f(5)-f(1)}{5-1}=\frac{2-0}{4}=\frac{1}{2}\]

Answer 4

so far so good?

Answer 5

yup

Answer 6

k, now the next thing you need is the derivative of \(f(x)=\sqrt{x-1}\)
you have that?

Answer 7

no but i can

Answer 8

you should do it in your head
let me know when you get it and what method you used

Answer 9

1/2(x-1)^-1/2 chain rule lol

Answer 10

ok good, but it would be better not to use exponential notation, because you are going to have to evaluate that thing

Answer 11

but thats how i leaned how to derrive chain rules lol show me what you mean

Answer 12

here is what i recommend
the square root is a very common function, so i would just remember that the derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\)

Answer 13

ok makes sence

Answer 14

it never changes, so there is no reason to always rewrite the square root as \(x^{\frac{1}{2}}\) and the use the power rule and then convert back

Answer 15

ok

Answer 16

so your derivative is, by the chain rule
\[\frac{1}{2\sqrt{x-1}}\]

Answer 17

it is what you had in any case, but now it is in a usable form
your last job is to solve
\[\frac{1}{2\sqrt{x-1}}=\frac{1}{2}\] for \(x\) and you are done

Answer 18

wait 1/2 isnt one of my answers

Answer 19

\(\frac{1}{2}\) is not the solution
you have to solve this equation
\[\frac{1}{2\sqrt{x-1}}=\frac{1}{2}\]

Answer 20

why did you set it = 1/2?

Answer 21

the mean value theorem says that if blah blah blah there is a number \(c\in (a,b)\) with
\[f'(c)=\frac{f(b)-f(a)}{b-a}\]
here we have
\[f'(c)=\frac{1}{2\sqrt{c}}\] and \[\frac{f(5)-f(1)}{5-1}=\frac{2-0}{4}=\frac{1}{2}\]

Answer 22

damn typo
i meant
\[f'(c)=\frac{1}{2\sqrt{c-1}}\]

Answer 23

that is why i we set it to \(\frac{1}{2}\) because that is the number we got for
\[\frac{f(b)-f(a)}{b-a}\] for \(a=1,b=5,f(x)=\sqrt{x-1}\)

Answer 24

ok i get but how do i solve for x ?

Answer 25

i would multiply both sides by 2 first

Answer 26

can you help me not sure how to start

Answer 27

\[\frac{1}{2\sqrt{x-1}}=\frac{1}{2}\\
\frac{1}{\sqrt{x-2}}=1\]

Answer 28

that means \(\sqrt{x-1}=1\)
oh another typo sorry
ignore the 2

Answer 29

x=0

Answer 30

x=2

Answer 31

\[\frac{1}{2\sqrt{x-1}}=\frac{1}{2}\\
\frac{1}{\sqrt{x-1}}=1\\
\sqrt{x-1}=1\] etc

Answer 32

yeah, 2

Answer 33

so the answer to my question is b) 2 only ????

Answer 34

yes

Answer 35

thank you satellite :) i think i've seen you in google lol. I remember satellite and purple. THank you <3

Answer 36

google? really?

Answer 37

yup lol

Answer 38

did you google "satellite73"? i would guess you got pictures of cars

Answer 39

no it was a math problem and you were the instructor lol

Answer 40

lol

Answer 41

ok thanks no i must read hamlet for english lol bye

Answer 42

hamlet more fun than math
enjoy