Justification with the mean value theorem: equation | AP Calculus AB | Khan Academy

- [Tutor] Let g of x equal one over x.

Can we use the mean value theorem

to say that the equation g
prime of x is equal to one half

has a solution where negative one

is less than x is less than two,

if so, write a justification.

Alright, pause this video

and see if you can figure that out.

So the key to using
the mean value theorem,

even before you even think about using it,

you have to make sure

that you are continuous
over the closed interval

and differentiable over the open interval,

so this is the open interval here

and then the closed interval
would include the end points.

But you might immediately realize

that both of these intervals
contain x equals zero

and if x equals zero,
the function is undefined

and if it's undefined there,

well it's not going to continuous

or differentiable at that point

and so no,

not continuous

or differentiable,

differentiable

over the interval,

over the interval.

Alright, let's do the second part.

Can we use the mean value theorem

to say that there is a value
c such that g prime of c

is equal to negative one half

and one is less than c is less than two

if so, write a justification.

So pause the video again.

Alright, so in this
situation, between one and two

on both the open and the closed intervals,

well, this is a rational function

and a rational function
is going to be continuous

and differentiable at
every point in its domain

and its domain completely contains

this open and closed interval

or another way to think about it,

every point on this open interval

and on the closed
interval is in the domain,

so we can write g of x

is a rational

function,

which lets us know that it is continuous

and differentiable at
every point in its domain,

at every point

in its domain,

the closed interval from one to two

is in domain

and so now, let's see what
the average rate of change is

from one to two

and so we get g of two

minus g of one

over two minus one is equal to one half

minus one over one, which is
equal to negative one half,

therefore,

therefore

by the

mean value theorem,

there must be a c

where one is less than c is less than two

and g prime of c is equal to

the average rate of change
between the end points,

negative one half

and we're done, so we can put
a big yes right over there

and then this is our justification.