Question

Determine the values m, M, a, b, and c such that the function f(x) satisfies the hypothesis of the MVT on the interval [-4,2]. sketch a graph of f.


mx+a; when -4 14 years ago

Answer 1

i know that the MVT says if f on [a,b] is continuous and differentiable, then there is an x=c within (a,b) such that f'(c)= f(b)-f(a) / b-a

Answer 2

f'(c) = mx+a (first condition)
a = -4, b = 2 (interval a to b)
f(b)-f(a) / 2+4 = mx-4
f(b)-f(a)/6 = mx - 4

Answer 3

or f(b)-f(a) = 6(mx - 4)
hmm im kinda confused now

Answer 4

i know the feeling

Answer 5

do you know anyone on the site who's good at this kind of problem?

Answer 6

@Zarkon @TuringTest can u guys take a look?

Answer 7

I assume that is supposed to be ,x+a when \[-4\le x\le 2\]?
(you have -2)

Answer 8

mx+a *

Answer 9

the negative 2 is correct

Answer 10

Oh I misread sorry

Answer 11

I get it now...

Answer 12

this is more tedious than hard.

Answer 13

you need to make f continuous and differentiable.

Answer 14

looks like c and b need to be zero

Answer 15

why? because they aren't (-)?

Answer 16

to be continuous you need \[\lim_{x\to 0}f(x)=f(0)=0\]

Answer 17

\[\lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}(Mx+c)=M\cdot 0+c\]
this must be 0 thus \(c=0\)

Answer 18

ok. but then how do you find M? it could be anything in the interval, right?

Answer 19

you get M from the differentiability condition.

Answer 20

what does that mean?

Answer 21

\[f'(0)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]
must exist

Answer 22

\[f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}\]

Answer 23

does M=2 ?
(I'm trying to follow along as well..)

Answer 24

looking at the limit as h goes to zero from the right

\[\lim_{h\to0^+}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0^+}\frac{Mh-0}{h}=M\]

Answer 25

\[\lim_{h\to0^-}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0^+}\frac{h^2+2h-0}{h}=2\]

thus, as TT stated, M=2

Answer 26

ok, I need to start from a beginning. I'm lost as to how we got there.

so if f(x)=x^2+2x+b when -2<x<0
first, i find f'(x) in the interval?

f'(x)=2x+2, right?
f'(x)=0 when x=-1?

Answer 27

for get a - sign on the limit

\[\lim_{h\to0^-}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0^-}\frac{h^2+2h-0}{h}=2\]

Answer 28

Thank you Zarkon for yet another brilliant explanation :D
(sorry if I interrupted it, I just wanted to make sure I understood)

Answer 29

not at all.

Answer 30

it's all good TT :)

Answer 31

I'm sorry, i looked at it again, and I'm more confused. I get how you find M, but not c

Answer 32

what is the limit of f(x) as x goes to 0 from the right?

Answer 33

0?

Answer 34

and as x goes to zero from the left ?

Answer 35

no holes, so it's 0, right?

Answer 36

sorry no it's c

Answer 37

so, is a=0 as well?

Answer 38

remember what a typed above...

\[\lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}(Mx+c)=M\cdot 0+c=c\]

Answer 39

Oook! so c is the limit.

Answer 40

I messed up zarkons explanation, let me try again...

Answer 41

what is the limit of f as x goes to zero from the left?
(that is when f(x)=x^2+2x+b)

Answer 42

is it b? or is it f'(x)? =2x+2?

Answer 43

right, and the value of f(0) is defined to be...?

Answer 44

b

Answer 45

it's b...

Answer 46

no f(0) is defined to be zero, yes?

Answer 47

if x=0
f(0)=0^2 + (2*0) + b

right? how do you get b=0?

Answer 48

so, i need to find where f'(x)=0 for each condition?

find all crotocal numbers for each interval?

Answer 49

**critical**

Answer 50

\[\lim_{x\to0^-}f(x)=b\]\[\lim_{x\to0^+}f(x)=c\]\[f(0)=0\]By definition thefunction will only be continuous at zero if\[\lim_{x\to0}f(x)=f(0)\]hence this is only continuous when\[\lim_{x\to0^-}f(x)=b=\lim_{x\to0^+}f(x)=c=f(0)=0\]

Answer 51

ok. that makes sense!

Answer 52

awesome, happy to help
all credit to Zarkon though; this would've taken me ages to figure out alone!

Answer 53

limx→0 f(x)=a

Answer 54

is a the same?

Answer 55

are we talking about the same problem?
if so, then no...

Answer 56

hmm. yea, i was talking about the same one.

how do you find the lim of a?

Answer 57

you need to look at f(-2) and \(\lim_{x\to -2^{-}}f(x)\)

Answer 58

lim f(x) = lim(mx+a) = m⋅0+a=a
x→0+ x→0+

Answer 59

right?

except with a negative. not a positive in the x-->0-

Answer 60

no, f is not defined that way for x around 0

Answer 61

we are not on the same page here...look at what i typed. above

Answer 62

right, what is f defined as in the interval we are talking about?

Answer 63

lim f(x) = lim f(mx+a) = m(-2) + a
x→−2(-) x→−2(-)

Answer 64

f(x)=mx+a in the interval, yes?

Answer 65

from the left, yeah

Answer 66

but the limit is -2, not 0. i got confused about the 0 because we just finished learning about critical #s and I keeo trying t set everything = 0

Answer 67

**keep

Answer 68

you always gotta look at the value that x is said to be approaching in the limit
it could be anythin;
x can go to infinity, zero, whatever

Answer 69

so, we have f(x)=-2m + a and lim x->0 for f(x) is -2

Answer 70

we find m using the definition of a derivative

Answer 71

like you did with M

Answer 72

but how do you get a?

Answer 73

you will need to find m first

Answer 74

derivative trick again

Answer 75

I though he had found m

Answer 76

he found M not m

Answer 77

oh right...

Answer 78

poor choice of variables :)

Answer 79

so yeah, derivative trick again
f'(-2) from the right must equal f'(-2) on the left

Answer 80

how do you mean?

Answer 81

same thing we did to find M
(another reason it confused me)

Answer 82

read above Zarkon's method if you didn't catch it

Answer 83

yea, i'm just wndering if it'd be [f(m+h) - f(m)] / h

Answer 84

it's always[f(x+h) - f(x)] / h

Answer 85

do you sub the whole mx+a in for x here?

Answer 86

f(x+h)=m(x+h)+a
gotta remember these kinds of things ;)

Answer 87

understanding them the first time around would've helped too, i guess...

Answer 88

yeah, lol
sometimes they don't teach first principles well :(
anyway think you can get M ?

Answer 89

so, you don't sub the function into the formula, you sub the formula into the function?

Answer 90

(m(x+h)+a - mx+a) / h ?

Answer 91

(m(x+h)+a - (mx+a) / h

Answer 92

(mx+mh+a-mx-a) / h
=mh / h

Answer 93

would it be -2=mh / h?

Answer 94

we are talking about M, yes?

Answer 95

m=-2

Answer 96

little m

Answer 97

in the first condition

Answer 98

oh right, so what about big M ?