CalculusAB AP test. How do I do this?

[DyslexicHobo]

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!

 

[nate808]

First, let's 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<r<3. So now, we plug in values 2-4 on the f(x), and we see that because it is continuos, at some point on that interval f(x) must equal 1. We need it to equal 1 so that it will satisfy the given h(r)=-5.

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.

part c) By definition, the derivative of that integral is just f(g(3)).

 

[pakmingki]

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 got to 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)

 

[DyslexicHobo]

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).

 

[phoenixthoth]

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?

 

[DyslexicHobo]

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)

http://users.adelphia.net/~sismondo/ (link to homepage)

 

[pakmingki]

if a function is differentiable on a given interval, that means it's derivative is continuous on that interval.

 

[phoenixthoth]

if a function is differentiable on a given interval, that means it's derivative is continuous on that interval.

Not true. For example:
$$f\left( x\right) =\left\{ \begin{array}{cc} 0, & x=0 \\ x^{2}\sin \left( 1/x\right) , & x\neq 0 \end{array} \right.$$

 

[mathwonk]

anyody can ask a question so it is hard to understand. I am a professional mathematician, and that **** table of values is so eye boggling i cannot even bring myself to read it. don't feel bad. these are just stupid standardized tests. In real life, the point is not to answer tricky questions, but to ponder hard ones.