yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

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


2m read
·Nov 11, 2024

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.

All right, 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 endpoints. But you might immediately realize that both of these intervals contain x equals 0, and at x equals 0 the function is undefined. And if it's undefined there, well, it's not going to be continuous or differentiable at that point.

And so no, not continuous or differentiable over the interval.

All right, 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.

All right, so in this situation between 1 and 2 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 this domain, at every point in its domain. The closed interval from 1 to 2 is in domain.

And so now let's see what the average rate of change is from 1 to 2.

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, 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 endpoints, negative one half.

And we're done. So we could put a big yes right over there, and then this is our justification.

More Articles

View All
Tech's Impact On Young Brains | America Inside Out with Katie Couric
As more young people like David pull up in their rooms with their devices, studies show a generation delaying adulthood. Fewer get driver’s licenses, have after-school jobs, or date. But most alarming, the suicide rate for girls ages 15 to 19 doubled betw…
Watch: What It’s Like to Read Lips | Short Film Showcase
So, when I was really young, probably kindergarten or first grade, I have a much older brother, and we’d go out to recess. There was this older guy; he might have been in like fifth or sixth grade. They’d always used to pick on us, and I didn’t really kno…
Michael Burry’s New Warning for the 2023 Recession
Michael Berry made his name betting against the housing market. It took two years for the drama to play out, but the subprime mortgage market finally collapsed in 2007, just as he had predicted. So, he made a ton of money, much more than I ever imagined I…
Why you don't have enough money
So pretend you’re this guy and you’re in bed typing in random country names on Google Flights, checking the prices because you know after the pandemic is over, you’re gonna travel the world and see and taste things you’ve never seen or tasted before. But …
Analyzing motion problems: total distance traveled | AP Calculus AB | Khan Academy
Alexi received the following problem: a particle moves in a straight line with velocity v of t is equal to negative t squared plus 8 meters per second, where t is time in seconds. At t is equal to 2, the particle’s distance from the starting point was 5 m…
Halloween and Neil deGrasse Tyson | StarTalk
I was never big into Halloween costumes. When I was a child, I had a costume, but I didn’t have so much invested in what it was or what it looked like that it became a part of my childhood memories. I grew up; my formative years were in a huge apartment …