# CalculusAB AP test. How do I do this?

by DyslexicHobo
Tags: calculusab, test
 P: 248 I took the Calculus AB AP test, and there were a total of 6 open-ended problems. Most of them I knew exactly how to do, but there was one in particular that gave me some real problems. We haven't done much on function composition, so maybe that's why I couldn't do it (but I have the feeling that I'm just overlooking something). This table is given: x f(x) f'(x) g(x) g'(x) 1 6 4 2 5 2 9 2 3 1 3 10 -4 4 2 4 -1 3 6 7 The functions f and g are differentiable for all real numbers, and g is strictly increasing. The table above gives the values of the functions and their first derivatives at selected values of x. The function h is given by h(x) = f(g(x)) - 6. a) Explain why there must be a value r for 1 < r < 3 such that h(r) = -5 b) Explain why there must be a value c for 1 < c < 3 such that h'(c) = -5 c) Let w be the function given by w(x) = integral of f(t)dt from 1 to g(x). Find the value of w'(3). d) If g^-1 is the inverse function of g, write an equation for the line tangent to the graph of y = g^-1(x) at x=2 So far, all I've established is that f and g are continuous, g'(x) will always be positive, and g(x) will always be less than g(x+1). Everything I know is just the obvious, I don't know where to go next! Can someone please give me the solution so I can quit worrying about this stupid problem? Thanks for any help!
 P: 56 First, lets take a look at part a). When you have a composite function f(g(x)), whatever the x value is (or r value in this case), you plug that number in for g(x), whatever value that yields, you then plug into f(x). So for part a, the x values are all increasing from (2,4) on the r interval 1
 P: 93 i agree with nate on most of the parts, but the explanation i used for b is as follows, and i think this is the explanation the graders are looking for. the average rate of change of h on (1,3) is -5. h(3) = -7, h(1) = 3 -7-3/2 = -5 Therefore, by the mean value theorem, there must be some value in (1,3) whose derivative is -5 because there has to be a value where the derivative is equal to the avg rate of change. for c, nate forgot the chain rule. w'(x) is actually f(g(x)) * g'(x). You gotta do the chain rule when endpoints are functions. So w'(3) is f(g(3)) * g'(3) for d, u must remember the equality that the derivative of the inverse of the y value is equal to 1/(derivative at x) so, it asks for the tangent line of the inverse of g when the function is at 2. Remembr with inverse, the ordered pairs are switched. So in the original function, it's the y value that is 2. So the ordered pair you will use is (2,1). So, the derivative is 1/(g'(1)) = 1/5 so the slope of the tangent line is y-1 = 1/5 (x-2)
P: 248

## CalculusAB AP test. How do I do this?

Thank you for the explanation. I feel kind of dumb for not knowing how to do this on the test. I made it seem much harder than it really was (I wasn't even able to get part of it).
P: 1,572
 Quote by nate808 part b) the derivative for a composite is defined as such if h(x)=f(g(x)), then h'(x)=f'(g(x))g'(x). (BTW the 6 cancels b/c it is a constant) As stated previoulsy, the g(x) values will be between 2-4, so now we must look at the f'(x) values on that interval. By plugging in values, you see that h'(2)=2 and h'(3)=-8. Because of continuity, it must =-5 on that interval.
You appear to be using the intermediate value theorem on h'. How do you know that h' is continuous?
P: 248
 Quote by phoenixthoth You appear to be using the intermediate value theorem on h'. How do you know that h' is continuous?
That's a good point! Can someone please prove that h'(r) is continuous?

Also, if anyone is interested, I found this site: http://users.adelphia.net/~sismondo/AB073.html (link to this problem)

$$f\left( x\right) =\left\{ \begin{array}{cc} 0, & x=0 \\ x^{2}\sin \left( 1/x\right) , & x\neq 0 \end{array} \right.$$