Justifying use of the mean value theorem: equation
About this video
For g(x) = 1/x, this video decides when the mean value theorem can be used. On -1 < x < 2 it cannot, because g is undefined at x = 0, so it is not continuous or differentiable there. On [1, 2], g is a rational function defined on the whole interval, the average rate of change is (1/2 - 1)/(2 - 1) = -1/2, so the theorem guarantees a c in (1, 2) with g'(c) = -1/2.
You'll be able to
- check whether a function is continuous and differentiable on an interval before using the mean value theorem.
- explain why a point where a function is undefined blocks the theorem.
- write a full mean value theorem justification for a rational function.
Key words
- rational function
- A fraction made of polynomials, such as 1/x; it is continuous and differentiable at every point in its domain.
- domain
- The set of x values where the function is defined; 1/x is not defined at x = 0.
- average rate of change
- (g(b) - g(a))/(b - a), the slope of the secant line.
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.
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