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### Course: AP®︎/College Calculus AB > Unit 9

Lesson 2: AP Calculus AB 2015 free response- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c

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# 2015 AP Calculus 2c

Rate of change of difference between functions.

## Want to join the conversation?

- should it be the absolute value, because the vertical distance can't be negative?(4 votes)
- Just by eyeballing it, we know f(x) - g(x) can't be negative because we're looking at region S, where f(x) > g(x)! So an absolute value function would be redundant.

If you're referring to the derivative, remember it can be negative because we're not talking about distances; we're talking about rates, and rates can be positive (increasing) or negative (decreasing).(3 votes)

- At0:43, how do I know it is h'(1.8)? It makes sense when Sal says it, but on a real AP test, how do I interpret the written question correctly to assume they are asking for the derivative (h'(x))?

Hope this question makes sense, and thanks so much in advance!(2 votes)- The question asked for the rate at which h(x) changes at x = 1.8. When it asks for the RATE at which something is changing, it is asking for the derivative. Did this answer your question?(4 votes)

- At0:43, how do I know it is h'(1.8)? It makes sense when Sal says it, but on a real AP test, how do I interpret the written question correctly to assume they are asking for the derivative (h'(x))?

Hope this question makes sense, and thanks so much in advance!(2 votes)- Well if you look at the wording it says "find the rate" so immediately you know it's the derivative with respect to X = 1.8. Always remember rate = derivative(3 votes)

## Video transcript

- [Voiceover] Let h be
the vertical distance between the graphs of
f and g and region s. Find the rate at which h
changes with respect to x when x is equal to 1.8. We have region s right over here. You can't see it that
well since I drew over it. And what you see when we're in region s, the function f is greater
than the function g. It's above the function g. We can write h of x, we could write h of x, as being equal to f of x minus g of x and what we wanna do is we wanna find the
rate at which h changes with respect to x. We could write that as h prime of, we could say h prime of x but we want the rate
when x is equal to 1.8. So h prime of 1.8 is
what we wanna figure out. We could evaluate f prime of 1.8 and g prime of 1.8 and to do that we would take the derivatives of each of these things and we
know how to do that. It's within our capabilities,
but it's important to realize when you're taking the AP test that you have a calculator
at your disposal. And a calculator can numerically integrate and can numerically evaluate derivatives. And so when they actually want us to find the area or evaluate an integral, where they give the endpoints, or evaluate a derivative at a point, well that's a pretty good
sign that you could probably use your calculator here. And what's extra good about this is we have already
essentially inputted h of x in the previous steps
and really in part A. I defined this function here, and this function is essentially h of x. I took the absolute value of it so it's always positive
over either region, but I could delete the
absolute value if we want. Delete that absolute value, then I have to get rid of
that parentheses at the end. Let me delete that. And so notice this is h of x. We have our f of x, which was 1 plus x plus e
to the x squared minus 2x, and then from that we subtract g of x. G of x was a positive x to the
fourth but we're subtracting so negative x to the fourth, let me show you g of x right over here. G of x is right over here, and notice we are subtracting it. So the y1, as I've defined
it in my calculator, and I just pressed this y
equals button right over here, that is my h of x. Now I can go back to the other screen, and I can evaluate its derivative
when x is equal to 1.8. I go to math. I scroll down. We have n derivative right here and so click Enter there. And then what are we gonna
take the derivative of? Well the function y sub 1
that I've already defined in my calculator. I can go to variables, y variables, it's already selected function
so I'll just press Enter. And I'll select the function y sub 1 that I've already defined. So I'm taking the derivative of y sub 1. I'm taking the derivative
with respect to x. And I'm going to evaluate that derivative, when x is equal to 1.8. That simple. And then I click Enter. And there you have it. It's approximately -3.812. So approximately -3.812. And we're done. One thing you might appreciate
from this entire question, and even the question 1, is that they really wanted to make sure that you understand the
underlying conceptual ideas behind derivatives and integrals. And if you understand
the conceptual ideas, of how do you use them to solve problems, and you have your calculator at disposal, then these are not too hairy, that these can be done fairly quickly.