# CalculusAB AP test. How do I do this?

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:
Code:
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!

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<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)).

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)

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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).

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?

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

if a function is differentiable on a given interval, that means it's derivative is continuous on that interval.
Not true. For example:
$$f\left( x\right) =\left\{ \begin{array}{cc} 0, & x=0 \\ x^{2}\sin \left( 1/x\right) , & x\neq 0 \end{array} \right.$$

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mathwonk