ARC and IRC or Derivative?

[hopelynnwelch]

So I have this practice problem here and I know how to calculate the ARC for interval [2,4] It's 1/2

But how do I find the IRC or the derivative at 2?

I know the formula for IRC but I am wondering how I know what value of h to use in the formula when just looking at a graph. I have done a couple of derivatives and all the values are finding the limit as x goes to 0 and I use a really small number like .0001. Is 2 my h value in this case?

I know the answer is A because it is in the book. But I hate multiple choice and want to know that I am not just getting lucky with figuring this out but that I am actually doing it right.

 

[Steven G]

View attachment 4875



But how do I find the IRC or the derivative at 2?

Compute the slope of the tangent line from the left and right of x= 2. Is the slope the same? So what is f'(2)?

 

[hopelynnwelch]

Compute the slope of the tangent line from the left and right of x= 2. Is the slope the same? So what is f'(2)?

Zero?

I have only calculated them prior knowing that

IRC = The Limit as x goes to 0 of [f(a+h) - F(a)] / h

I think I can look at it and tell it is zero but I want to plug some numbers in and prove but I am not sure what I'm looking at or what to plug if that makes sense... lol

 

[Steven G]

Zero?

I have only calculated them prior knowing that

IRC = The Limit as x goes to 0 of [f(a+h) - F(a)] / h

I think I can look at it and tell it is zero but I want to plug some numbers in and prove but I am not sure what I'm looking at or what to plug if that makes sense... lol

Yes, it is zero or surely looks it. You can only use The Limit as x goes to 0 of [f(a+h) - F(a)] / h if you know what f(x) is and you don't. So you have to approximate. How about we agree that f ' (2) is close to 0? This is enough to answer the questions.

 

[hopelynnwelch]

Yes, it is zero or surely looks it. You can only use The Limit as x goes to 0 of [f(a+h) - F(a)] / h if you know what f(x) is and you don't. So you have to approximate. How about we agree that f ' (2) is close to 0? This is enough to answer the questions.

Okay. I can live with that.

Thank you!

 

[stapel]

View attachment 4875

So I have this practice problem here and I know how to calculate the ARC for interval [2,4] It's 1/2

I will assume that "ARC" means "average rate of change". Also, I will assume that you are asking for help on part (a) of exercise #35.

If you are computing the average rate of change over the specified interval as being the slope between the endpoints, then your value appears (at least approximately -- who can tell, exactly, from the picture, right?) to be correct.

But how do I find the IRC or the derivative at 2?

I will assume that "IRC" means "instantaneous rate of change" or, which is the same thing, the derivative. From another of your posts, it appears that you are working with the very beginnings of derivatives, so that you're still using intuitive concepts and limit definitions. (This is fine. It's just that it limits the tools at our disposal.)

I am wondering how I know what value of h to use in the formula when just looking at a graph.

You have to guess, based on the pretty picture and what you know (from algebra) about slopes. The fact that, right at x = 2, the graph looks really horizontal suggests that the slope (at x = 2) is going to be equal to, or really close to, m = 0. Then the derivative (what I think you're calling the IRC) is also going to be equal to, or really close to, zero.

For future reference, when you're guessing from a picture, don't be shy about grabbing a ruler, drawing a line that looks reasonable to you, and estimating a slope from that. You may find it helpful to think of the line as the surface of the snow-covered ground, "the point" as the place a skier is standing (as he moves across the graph to the right), and the derivative at the point being the slope of the straight line formed by the skier's skis as he's standing there. In this case, at x = 2, he seems to be standing on a fairly flat spot, so the slope is going to be zero. :wink: