Question

f(x)=1/x
Find all c guaranteed to exist by means value theorm. on interval [1,3]
i know the derivative is 1/x^2
and the slope of the function is 1/2
i set the derivative equal to 1/2 to find c but i get
1=-4/3x^2
i cant take the square root of a negative. can someone help me out on what im doing wrong or what i can do?

Answer 1

\[\Large\rm f(x)=\frac{1}{x}\]So our derivative,\[\Large\rm f'(x)=-\frac{1}{x^2}\]There should be a negative in there, ya?

Answer 2

And I'm not sure about that slope, let's check..

Answer 3

\[\Large\rm m=\frac{y_2-y_1}{x_2-x_1}=\frac{\frac{1}{3}-\frac{1}{1}}{3-1}=\frac{-\frac{2}{3}}{2}=-\frac{1}{3}\]

Answer 4

Eggg where you gooo >.< Bahhh you left lol

Answer 5

no lol i was starting to do chemistry lol. and are you sure the slope is not-4/3 because when we divide by 2 the reciprocal is multiplied i though? @zepdrix

Answer 6

Maybe I made a boo boo, lemme check real quick.

Answer 7

\[\Large\rm \frac{-\frac{2}{3}}{(2)}=\frac{-\frac{2}{3}}{(\frac{2}{1})}=-\frac{2}{3}\cdot\frac{1}{2}=-\frac{1}{3}\]Make sure you flip the 2 correctly.
Something like that, yah? :)

Answer 8

oh yeah im wrong lol . do you think +/-sqrt(3) is a good answer.

Answer 9

help

Answer 10

\[\Large\rm x=\pm\sqrt{3}\]So then,\[\Large\rm x=\sqrt3\]\[\Large\rm x=-\sqrt3\]Hmmm one of these values does NOT lie in our interval, so you can ignore it.

Answer 11

so the answer is sqrt(3) and the not the neative does hat sound right?!

Answer 12

Yes!!!

Answer 13

omg yay ^.^ but to use the Mean Value Theorm I have to verify it is possible on this question.
I say it is because
The domain is all real numbers except zero so it is continious on the closed interval.
And also it is differentiable on the open interval because the domain of the derivative is continuous on the open interval.
is that correct or am i mixed up.