Mean value theorem problems

[John45]

I am having difficulty with these problems I am getting that the theorem does not apply to each but I feel that my answers are wrong...

Directions state "determine whether the mean value theorem applies to the following function on the given interval [a,b] , if so, find or approximate the points guaranteed to exist by the theorem"

1) f(x)= ln2x ; [1,e]


2) x / (x+2) ; [-1,2]


--------------------------------------------------------------
so for number 1 i got
f(a)~ .6931
f(b)~ 1.6931

f(b)-f(a) = -1

-1 / (b-a) is aprroximately -.581976069
now I believe the derivative is 1 / 2x and I set the above value euqal to the derivative and solve for x and get approximately -250/291 but this value is not in the interval so does that mean that the theorem does not apply?
-------------------------------------------------------------
the second one I got

f(a)=-1
f(b)= .5

f(b)-f(a)= -.5

-.5 / (b-a) = -.5

the derivative of x / (x+2) I believe is 2 / (x+2)^2 I set it equal to -.5 and solve for x and get [Sqrt(2) + 1] / -.5 this is also not in the domain so I assume the theorem does not apply??
_________________________________________________
also lastly I must Find all functions of f whose derivative if f`(x)=x+1

I feel very stupid for asking that question because it seems so simple I know that the derivative of x is 1 for the +1 part but the x is losing me

Thanks for any assistance

 

[Deleted member 4993]

I am having difficulty with these problems I am getting that the theorem does not apply to each but I feel that my answers are wrong...

Directions state "determine whether the mean value theorem applies to the following function on the given interval [a,b] , if so, find or approximate the points guaranteed to exist by the theorem"

1) f(x)= ln2x ; [1,e]


2) x / (x+2) ; [-1,2]


--------------------------------------------------------------
so for number 1 i got
f(a)~ .6931
f(b)~ 1.6931

f(b)-f(a) = 1.6931 - 0.6931 = 1


-1 / (b-a) is aprroximately -.581976069
now I believe the derivative is 1 / 2x --- Incorrect

$\displaystyle \frac{d}{dx}ln(2x) \ = \ \frac{2}{2x} \ = \ \frac{1}{x}$

and I set the above value euqal to the derivative and solve for x and get approximately -250/291 but this value is not in the interval so does that mean that the theorem does not apply?
-------------------------------------------------------------
the second one I got

f(a)=-1
f(b)= .5

f(b)-f(a)= -.5 <<< How's that !!! f(b) - f(a) = 0.5 - (-1) = 1.5
-.5 / (b-a) = -.5

the derivative of x / (x+2) I believe is 2 / (x+2)^2 I set it equal to -.5 and solve for x and get [Sqrt(2) + 1] / -.5 this is also not in the domain so I assume the theorem does not apply??
_________________________________________________
also lastly I must Find all functions of f whose derivative if f`(x)=x+1

I feel very stupid for asking that question because it seems so simple I know that the derivative of x is 1 for the +1 part but the x is losing me

Thanks for any assistance

 

[galactus]

Directions state "determine whether the mean value theorem applies to the following function on the given interval [a,b] , if so, find or approximate the points guaranteed to exist by the theorem"

Here's the first one to give you the idea.

1) f(x)= ln2x ; [1,e]

$\displaystyle f'(x)=\frac{1}{x}$

$\displaystyle f'(c)=\frac{1}{c}$

$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{ln(2)+1-ln(2)}{e-1}=\frac{1}{e-1}$

$\displaystyle f(a)=ln(2)$

$\displaystyle f(b)=ln(2\cdot e)=ln(2)+1$

$\displaystyle \frac{1}{c}=\frac{1}{e-1}\Rightarrow c=e-1$

By the MVT, the point in the interval [1,e] guaranteed by the theorem is $\displaystyle c=e-1\approx 1.718...$

There is a line tangent to the curve y=ln(x) somewhere in between 1 and 2 that has the same slope as the secant line joining $\displaystyle (1, \;\ ln(2))$ and $\displaystyle (e, \;\ ln(2)+1)$

This tangent line has equation $\displaystyle y=.582x+.111$

Note the slope is about .582. The secant line, f'(c), has slope $\displaystyle \frac{1}{e-1}\approx .582$

See what is going on?.