Main content
AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > Unit 5
Lesson 1: Using the mean value theorem- Mean value theorem
- Mean value theorem example: polynomial
- Mean value theorem example: square root function
- Using the mean value theorem
- Justification with the mean value theorem: table
- Justification with the mean value theorem: equation
- Establishing differentiability for MVT
- Justification with the mean value theorem
- Mean value theorem application
- Mean value theorem review
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Justification with the mean value theorem: equation
AP.CALC:
FUN‑1 (EU)
, FUN‑1.B (LO)
, FUN‑1.B.1 (EK)
Example justifying use of mean value theorem (where function is defined with an equation).
Want to join the conversation?
Can we get more explanation on1:16"All Rational functions are continuous and differentiable in it's domain"(7 votes)
A rational function is one polynomial divided by another. Polynomials are differentiable everywhere, so we can always differentiate a rational function using the quotient rule, unless the denominator is zero at that point.
But if the denominator is 0 at some point a, then a is not in the domain of the rational function. So for every point that is in the domain, we can differentiate the function normally.(9 votes)
- At1:16, if all rational functions are continuous and differentiable at every point in its domain, why is g(x) not continuous at x = 0? Isn't x = 0 a part of its domain?(2 votes)
The domain of a function is the set of input values for which the function is defined.
𝑔(𝑥) = 1∕𝑥 is not defined for 𝑥 = 0 (due to division by zero), and thereby 𝑥 = 0 is not part of its domain.(10 votes)
at2:13how do we find out that g(2) is 1/2?(2 votes)- The function is given to be g(x) = 1/x, so g(2) = 1/2(3 votes)
for the given information I solved it like this,
1.
g(x)=1/x
g'(x)=-1/x^2
given that,
g'(x)=1/2
-1/x^2=1/2
x=(-2)^(1/2) which is not the element of (-1,2)
therefore, No
2.
g(x)=1/x
g'(x)=-1/x^2
given that,
g'(c)=-1/2
-1/x^2=1/2
x=(2)^(1/2) which is the element of (1,2)
therefore, Yes
Is this okey?(3 votes)- Where does the equation at2:14come from and how did Sal know it was equal to fprime of c?(2 votes)
- Because that's what the Mean Value Theorem states: If a function f(x) is differentiable on an interval [a,b], then there is a point c in [a, b] where f'(c)=[f(b)-f(a)]/(b-a).
This is just an application of the mean value theorem with a=1, b=2.(0 votes)
- Why is differentiability necessary for the validity of the Mean Value Theorem?(0 votes)
If the function is not differentiable, (i.e. it is discontinuous and/or contains a sharp curve), we cannot guarantee that there exists a point in the interval whose derivative is equal to the slope of the line passing through the points (a,f(a)) and (b,f(b)).(1 vote)
- why do we write -1 < x < 2 or 1 < c < 2? we need the function to be continuous over the closed interval. shouldn't we write -1 <= x <= 2 or 1 <= c <= 2?(0 votes)
- what if the equation is y=e^x(0 votes)
It is continuous as well as differentiable for all x. So, MVT can be applied.(0 votes)
- Quick question (about part 2 in the video):
So if the f'(c)=-1/2, the only value of x (I can find) that makes that true is x= 4 which is outside our interval (1,2)...(0 votes)
If you take the derivative of g(x)=1/x, you get g'(x)=-1/(x^2). Solve the equation -1/(x^2)=(-1/2) for x and you get two possible solutions: + or - the square root of 2. The negative square root is not in the interval (1,2). The positive square root of 2, which is approx. 1.414 is in the interval (1,2).(1 vote)
1:16
Video transcript
- [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.